A 


Course  in  Descriptive  Geometry  and  Stereotomy, 

BY  S.   EDWARD   WARREN,   C.E., 

PROFESSOR  OP  DESCRIPTIVE    GEOMETRY,   ETC.,    IN    THE    MASS.    INSTITUTE    OF 

TECHNOLOGY,   BOSTON.     FORMERLY  IN  THE  RENSSELAER 

POLYTECHNIC   INSTITUTE,   TROY. 

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GENERAL   PROBLEMS 


SHADES    AND    SHADOWS. 


FORMED  BOTH  BY  PARALLEL  AND  BY  RADIAL  RAYS ;  AND 

SHOWN  BOTH  IN  COMMON  AND  IN  ISOMETRICAL 

PROJECTION:    TOGETHER    WITH    THE 

THEORY   OF    SHADING. 


S.   EDWAED    WARREN,   C.E., 

PBOF.  OP  DESCRIPTIVE  GEOMETRY,  ETC.,  IN  THE  RENSSELAER  POLYTECHNIC  INSTITUTE;  AND  AUTHOP 

OP  "ELEMENTARY  PLANE  PROBLEMS;"  "DRAFTING  INSTRUMENTS,  ETC.;"  "ELEMENTARY 

PROJECTION    DRAWING;"    "ELEMENTARY    LINEAR    PERSPECTIVE;" 

AND    "  DESCRIPTIVE    GEOMETRY." 


NEW  YORK: 
JOHN    WILEY    &    SON, 

15  ASTOR  PLACE. 
1875. 


Entered  according  to  Act  of  Congress,  in  the  year  186T,  by 
JOHN  WILEY  &  SON, 

In  the  Clerk's  Office  of  the  District  Court  of  the  United  States  for  the  So  them   District 

of  New  York. 


JOHN  F.  TROW  &  SON,  PRINTERS, 
805-313  EAST  i3TH  ST.,  NBW  YORK. 


CONTENTS. 


GENERAL  TABLE ix 

PREFACE xi 

BOOK  I. — Construction  of  Shades  and  Shadows 1 

PART   L — IN  COMMON  PROJECTION ,  1 

SERIES  L — By  parallel  rays 1 

General  Principles 1 

§  1. — Definitions  and  classification  of  problems 1 

§  2. —  Graphical  representation  of  Shades  and  Shadows 5 

DIVISION  I. — Shades  and  Shadows  on  Ruled  Surfaces 7 

CLASS!.—  "  "  PLANK         "        7 

§  L— Shades 7 

PROBLEM  I. — To  determine,  first  whether  the  line  of  intersection 
of  two  oblique  planes  is,  or  is  not,  a  line  of  shade ;  and, 
second,  to  determine,  by  inspection,  the  compound  line  of 
shade  on  a  given  plane  sided  body 8 

PROBLEM  II. — To  determine  the  edges  of  shade  on  a  pyramid, 
whose  axis  is  oblique  to  both  planes  of  projection. — Two 
methods 9 

§  II.— Shadows 11 

PROBLEM  IIL — To  find  the  shadow  of  a  straight  line  on  the  hori- 
zontal plane  of  projection 11 

PROBLEM  IV. — To  construct  the  shadow  of  a  square  abacus  upon  a 
square  pillar,  and  of  both,  on  the  horizontal  plane  of  projec- 
tion    12 

PROBLEM  Y. — Having  the  projections  of  a  ray  of  light,  upon  two 
vertical  planes  at  right  angles  to  each  other ;  to  find  its  projec- 
tions on  those  pknes,  after  revolving  it  90°  about  a  vertical 
axis 13 

PROBLEM  VI. — To  find  the  projections  of  the  shadows  on  the  parts 
of  a  timber  framing,  shown  in  two  elevations,  on  two  vertical 
planes  at  right  angles  to  each  other 14 


IV  CONTENTS. 

PAGB 

PROBLEM  VII. — To  construct  the  shadow  of  a  circle  on  the  vertical 

plane  of  projection 16 

PROBLEM  Till. — To  construct  the  shadow  of  a  straight  lii  e  on  a 

plane  whoso  traces  are  parallel  to  the  ground  line 19 

PBOBLESI  IX. — To  find  the  shadow  of  a  chimney,  situated  on  the 

end  portion  of  a  hipped  roof,  upon  the  end  and  side  roofs. .  19 
PROBLEM  X. — To  find  the  shadow  of  a  shelf  and  brackets  upon  a 

vertical  plane 22 


CLASS    H. — SHADES   AND   SHADOWS   ON   SINGLE   CCRVED    SURFACES,   IN 

GENERAL 23 

SECTION  L — On  Developable  Single  Curved  Surfaces 23 

§  I.— Shades 23 

PROBLEM  XI. — To  construct  the  elements  of  shade  on  any  cylinder 

or  cone 24 

First. — On  a  horizontal  right  semi-cylinder 24 

Second. — On  any  oblique  cylinder 24 

Third. — On  a  cylinder,  whose  axis  is  parallel  to  the 

horizontal  plane  only 24 

Fourth* — On  a  cone 25 

Fifth, — On  two  intersecting  cylinders 26 

§  H.— Shadows 27 

PROBLEM  XII.— To  construct  the  shadow  of  the  projecting  head  of 

a  cylinder  upon  the  cylinder 27 

PROBLEM  XIII. — To  construct  the  shadow  of  the  edges  of  shade 

of  a  cross  upon  a  cylinder,  both  bodies  being  seen  obliquely.       28 
PROBLEM  XIV. — To  construct  the  shadow  cast  by  the  upper  base 

of  a  hollow  oblique  cylinder  upon  its  interior. 30 

PROBLEM  XV. — To  construct  the  shadow  of  the  upper  base  of  a 
vertical  right  cone,  upon  the  lower  nappe  of  the  same 

cone 32 

PROBLEM  XVI. — To  construct  the  shadow  cast  by  the  vertical  cylin- 
der (Prob.  XI.,  Fifth)  upon  the  horizontal  cylinder,  and  by 

the  horizontal  cylinder  upon  the  vertical  cylinder 34 

1°. — Of  the  element  of  shade  of  the  horizontal 

cylinder  on  the  vertical  cylinder 35 

2°. — Of  the  upper  base  of  the  vertical  cylinder 

upon  the  horizontal  cylinder 36 

8°. — Of  the  element  of  shade  of  the  vertical  cylin- 
der upon  the  horizontal  cylinder 36 

PROBLEM  XVII. — To  find  the  axes  of  an  elliptical  shadow,  two  of 

whose  conjugate  diameters  are  known 37 

PROBLEM  XVIII. — Having  a  vertical  semi-cylinder  with  its  meridian 
plane  parallel  to  the  vertical  plane  of  projection,  it  is  required 


CONTENTS.  V 

PACK 

to  find  the  shadow  cast  on  its  base  and  visible  interior,  by 
its  edge  of  shade,  and  by  a  vertical  semicircle,  described  on 

the  diameter  of  its  upper  base 3C 

1°. — The  shadow  of  the  vertical  edge 39 

2°. —             "            "    front  semicircle 39 

3°. —             "      on  the  upper  base  circle 39 

4°. —             "      by  auxiliary  spheres 39 


SECTION  IL — Shades  and  Shadows  on  Warped  Surfaces 43 

§  L— Shades 41 

PHOBLEM  XIX. — To  construct  the  curve  of  shade  on  a  hyperbolic 

paraboloid  42 

PHOBLEM  XX. — To  find  the  curve  of  shade  on  the  common  oblique 
helicoid,  in  the  practical  case  of  the  threads  of  a  triangular- 
threaded  screw 45 

1°. — Description  of  the  screw 45 

2°. — To  construct  any  particular  element.     (Two 

methods.) 46 

Points  of  shade  on  assumed  helices 46 

First. — By  planes  of  given  declivity. , 4*7 

1°. — On  an  outer  helix. 
2°._     "      inner    " 

Second. — By  helical  translation 48 

Points  of  shade  on  assumed  elements.    Discussion 53 

PROBLKM  XXI. — To  construct  points  of  the  curve  of  shade  of  a 

conoid.     Discussion 55 

§  II.— Shadows 58 

PROBLEM  XXII. — To  find  the  several  shadows  cast  by  a  triangular- 
threaded  screw  upon  itself 60 

First.— Of  the  nut 60 

1st. — Of  any  unknown  point  of  an  edge 60 

2d. — Of  any  assumed  point  of  an  edge 62 

Second. — Of  the  curves  of  shade 63 

1st. — On  an  assumed  element. 63 

2d. — On  an  assumed  helix 64 

3d. — On  another  branch  of  the  curve  of  shade ....  64 

Third. — Of  the  outer  helix 64 

1st. — On  an  assumed  elemont 64 

2d. — On  any  helix 65 

3d. — On  particular  elements 65 

Fourth. — The  shadow  on  the  horizontal  plane 66 

1st. — The  determination  of  the  lines  casting  this 

shadow 66 

•  2d. — The  Shadow  of  any  point 66 

PROBLEM  XXIII. — To  determine,  in  general,  the  shadows  of  a  ver- 
tical right  conoid 66 


VI  CONTENTS. 


PAGB 

DIVISION    n. — Shades    and    Shadows    on    Double    Curved 

Surfaces 68 

§  I. — Shades 68 

PROBLEM  XXIV.— To  find  the  curve  of  shade  of  the  sphere,  using 

only  one  projection  of  the  sphere 09 

PROBLEM  XXV. — To  find  the  curve  of  shade  on  an  ellipsoid  of  re- 
volution    11 

PROBLEM  XXVI. — To  find  the  projections  of  the  curve  of  shade  on 

an  ellipsoid  of  three  unequal  axes 72 

PROBLEM  XXVII. — To  construct  the  curve  of  shade  upon  a  torus.  75 

1°. — Four  points  on  the  visible  boundaries 75 

2°. — The  highest  and  lowest  points 75 

3°. — Intermediate  points 76 

PROBLEM  XXV1IL— To  find  the  curve  of  shade  on  a  picdouche. . .  76 

1°. — Preliminary  points  of  shade 77 

2°. — Intermediate  points,  by  tangent  cones 78 

3°. —                                  by  tangent  spheres 79 

§  II.— Shadows 81 

PROBLEM  XXTX. — To  construct  the  shadow  of  the  front  circle  of 

a  niche  upon  its  spherical  part 82 

1  °.— First  Solution.    General  Method 82 

2°. — Second  Solution.    Special  Method 83 

3°. — Point  where  the  shadow  leaves  the  spherical 

part , 84 

4°. — Third  Solution.     Special  Method.    Discussion. 

Various  constructions  of  e : 84 

PROBLEM  XXX.— To  find  the  shadow  of  the  upper  circle  of  a  pie- 
douche  upou  its  concave  surface 88 

1°. — The  highest  and  lowest  points  of  shadow. ...  88 

2°. — Any  intermediate  points.     Discussion 88 

§  III — General  Problem  in  Review  of  Shades  and  Shadows. 

determined  by  Parallel  Rays 91 

PROBLEM  XXXI. — To  find  the  shades  and  shadows  of  the  shaft 

and  base  of  a  Roman  Doric  Column 92 

PROBLEM  XXXII. — To  find  the  shades  and  shadows  of  the  capital 

and  shaft  of  a  Roman  Doric  Column 95 


SERIES  II. — Shades  and  Shadows  determined  by  diverging  rays  97 

SECTION  I. — General  Principles 97 

"      II. — Problems  involving  diverging  rays 99 

PROBLEM  XXXIIL — To    find    the   shadow  of  a   semi-cylindrical 

abacus,  upon  a  vertical  plane  through  its  axis 99 


CONTENTS.  Vll 

PAGE. 

PROBLEM  XXXIV. — To  find  the  curve  of  shade  on  a  sphere,  the 

light  proceeding  from  an  adjacent  point 100 

PBOBLEM  XXXV. — Having  a  niche,  whose  base  is  produced,  form- 
ing a  full  circle ;  and  a  right  cone,  the  centre  of  whose  cir- 
cular base  coincides  with  the  centre  of  the  base  of  the  niche, 
it  is  required  to  find  the  shades  and  shadows  of  this  system, 
when  illumined  by  an  adjacent  point 101 


PAET  IL — SHADES  AND  SHADOWS  IN  ISOMETRICAL  PROJECTION 103 

Section  L — General  principles 103 

PROBLEM  XXXVI. — To  find  the  angle  made  by  the  isometrical  ray 

of  light  with  the  isometrical  plane  of  projection 104. 

Section  n. — Shades  and  shadows  on  isometrical  planes 105 

PROBLEM  XXXV II. — Having  given  a  cube,  with  thin  plates  pro- 
jecting vertically  and  forward,  in  the  plane  of  its  left  hand 
back  face,  to  find  the  shadows  of  the  edges  of  these  plates 

upon  the  cube  and  its  base 105 

Section  TTT — Shades  and  shadows  on  non-isometrical  planes 106 

PROBLEM  XXXVni. — To  find  the  shadow  of  a  hexagonal  cupola 
on  a  coupled  roof,  one  face  of  the  cupola  making  equal  angles 

with  two  adjacent  walls  of  the  house 106 

Section  IV. — Shades  and  Shadows  on  single  curved  surfaces 107 

PROBLEM  XXXIX.— To  find  the  elements  of  shade  on  an  inverted 
hollow  right  cone,  the  shadow  on  its  interior,  and  the  shadow 

of  one  of  its  elements  of  shade  on  an  oblique  plane 107 

Section  V. — Shades  and  Shadows  on  double  curved  surfaces 108 

PROBLEM  XL. — To  find  the  curve  of  shade  on  a  sphere 108 

PROBLEM  XLI. — To  construct  the  curve  of  shade  on  a  torus 109 


BOOK  IL— The  finished  execution  of  shades  and  shadows 112 

CHAPTER  I. — THEORY  AND  CONSTRUCTION  OF  BRILLIANT  POINTS  ;  AND  OP 

GRADATIONS   OP   SHADE 112 

SECTION  I. — Preliminary  general  principles 112 

§  1°. — Geometrical  conditions  for  the  adequate  graphical  representa- 
tion of  form 112 

§  2°. — Of  the  physical  conditions  for  the  visibility  of  bodies,  and  an 

adequate  representation  of  their  forms 113 

§  3°.— Of  brilliant  points  and  lines 117 

SECTION  II. — The  construction  of  brilliant  points  and  lines 118 

§  1.— Brilliant  elements  of  planes 118 


Vlll  CONTENTS. 

PAGE 

PROBLEM  XLII. — To  find  the  brilliant  point  of  a  plane  which  re- 
ceives light  from  a  near  luminous  point 119 

§  2. — Brilliant  dements  on  developable  surfaces 119 

PROBLEM  XLIIL— To  find  the  brilliant  element  on  a  cylinder,  illu- 

•  minated  by  parallel  rays.     (Two  solutions). 119 

PROBLEM  XLIV. — To  find  the  brilliant  point  of  a  cylinder  which  is 

illuminated  by  diverging  rays 121 

PROBLEM  XLV. — To  find  the  brilliant  point  on  a  cone  which  is  illu- 
minated from  a  near  luminous  point 122 

§  3. — Brilliant  points  on  warped  surfaces 123 

PROBLEM  XL VI. — To  find  the  brilliant  element  of  a  warped  surface 

when  illuminated  by  parallel  rays r 123 

PROBLEM  XLVII. — To  construct  the  brilliant  point  on  a  screw, 

when  illuminated  by  parallel  rays 124 

§  4. — Brilliant  points  on  double  curved  surfaces 125 

PROBLEM  XLYIII. — To  find  tho  brilliant  point  on  any  double-curved 

surface,  when  illuminated  by  parallel  rays 125 

PROBLEM  XLIX. — To  find  the  brilliant  point  on  a  sphere,  illuminat- 
ed by  parallel  rays 126 

PROBLEM  L. — To  find  the  brilliant  point  on  a  piedouche 126 

CHAPTER  n. — OF  THE  REPRESENTATION  OP  THE  GRADATIONS  OP  LIGHT 

AND   SHADE..  127 


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PREFACE. 


WHEREFORE  thus  pursue  a  shadow,  some  may  ask,  seeing  the  title  of 
this  volume,  and  the  complexity  of  some  of  its  figures. 

We  will  endeavor  to  reply. 

The  study  of  Shades  and  Shadows  is  an  application  of  the  general 
problems  of  Descriptive  Geometry,  in  connexion  with  a  few  physical 
principles ;  and,  of  Descriptive  Geometry,  no  one,  who  has  occasion  to 
be  conversant  with  forms,  singly  or  in  combination,  can  know  too 
much,  either  for  practical  purposes,  or  as  a  promoter,  in  its  peculiar 
way,  of  mental  power. 

In  particular,  having  in  a  previous  work  on  the  General  Problems  of 
Descriptive  Geometry,  disposed  of  the  Projections,  Intersections,  Tan- 
gencies,  and  Developments  of  geometrical  forms  in  the  abstract,  the 
way  is  prepared  for  the  extended  direct  application  of  the  second  and 
third  of  these  operations,  involving  incidentally  that  of  the  first,  to  the 
concrete  geometrical  subject  of  Shades  and  Shadows.  For  a  Shade  is 
always  bounded  by  a  line  of  contact,  and  a  Shadow  by  one  of  intersec- 
tion ;  and  before  either  can  be  found,  the  body  casting,  and  the  surface 
receiving  the  shadow,  must  be  given  by  their  projections. 

The  easy  logical  sequence  of  "  General  Descriptive "  Problems,  and 
"  Shades  and  Shadows "  as  a  theoretical  branch  of  applied  "  Descrip- 
tive," is,  therefore,  one  point  of  interest  attaching  to  the  study  of  the 
latter  subject. 

Further,  there  is  beauty  in  the  idea,  that  for  any  distance  and  direc- 
tion of  the  source  of  light,  and  for  any  form  and  position  of  the  bodies 
casting  and  receiving  shadows,  the  mind  can  know,  and  the  hand  can 
execute  those  shadows  truly. 

The  utility,  however,  of  delineated  shades  and  shadows,  in  rendering 
working  drawings  at  once  not  only  more  beautiful,  but  more  intelligible, 
because  more  nearly  conformed  to  reality,  is  the  chief  ground  of  inte- 


xil  PREFACE. 

rest  in  the  study  of  them.  The  student,  therefore,  who  has  experienced 
any  degree  of  pleasure  in  the  pursuit  of  the  general  problems  of 
"  Descriptive,"  may,  without  probable  disappointment,  promise  himself 
added  enjoyment  in  the  study  here  introduced  to  him. 

Once  more :  The  many  interesting  theoretical  and  practical  particu- 
lars in  which  the  science  of  optics  is  found  related  to  the  special  study 
of  shades  and  shadows,  lend  interest  to  the  latter,  as  affording  a  field 
for  observation  and  reflection  to  be  enjoyed  in  common  by  the  artist, 
the  physicist,  and  the  geometrical  draftsman.  Indeed,  it  is  hoped  that 
Book  II.  of  this  volume,  especially  Chapter  II.,  on  the  execution  of 
shades  and  shadows,  will  be  found  worthy  of  the  attention  of  artists. 

A  word  now  as  to  the  plan  of  this  work. 

Great  care  has  been  bestowed  on  the  statements  of  general  principles, 
which,  in  any  effective  study  of  the  subject,  must  be  mastered  as  fully 
as  the  problems. 

Again :  with  his  mind  properly  guarded  against  perplexity,  by  due 
comprehension  of  a  simple,  and  manifestly  rational,  classification  of 
the  problems  of  a  proposed  course,  the  student  will  gain  little  effective 
assistance  from  the  study  of  a  protracted  series  of  similar  problems. 

One  good  characteristic  problem,  under  each  of  the  heads  pointed 
out  in  the  preceding  general  table,  would  therefore  be,  at  least  in 
theory,  sufficient. 

The  student  having  mastered  these  representative  problems,  would 
then  readily  refer  any  new  problem,  met  by  him,  to  its  appropriate 
'group ;  and  would  then  solve  it  by  an  intelligent  use  of  the  general 
principles  applicable  to  all  problems  of  that  group. 

But  it  is  usually  difficult  for  the  beginner  to  dispossess  himself 
entirely  of  the  idea  that  problems,  which  differ  considerably  in  out- 
ward form,  do  not  differ  likewise  in  essential  principle,  and  method  of 
solution.  Besides,  there  are  often  indirect  and  special  methods  oi 
solution  which  really  need  separate  exemplification  ;  hence,  a  liberal 
number  of  special  problems,  some  of  them  of  quite  a  practical  cast, 
have  been  scattered  through  this  work. 

With  this  modification  of  what  was  at  first  pointed  out  as  a  theoreti- 
cally sufficient  course,  it  is  hoped  that  the  present  effort  will  be  accept- 
able to  professors  and  artists,  and  to  classes  in  scientific  collegiate 
courses,  and  in  the  Polytechnic  Schools ;  especially  to  those  who  have 
already  become  familiar  with  the  rudiments  of  Shades  and  Shadows,  as 


PREFACE.  Xlli 

presented  in  the  author's  "Elementary  Projection  Drawing,"  or  in 
similar  works. 

As  to  methods  of  study,  none  is  better,  after  reading  the  text  till  it 
is  fully  understood,  than  repeated  rehearsals  of  the  problems,  and  prin- 
ciples, by  the  student,  with  the  book  closed ;  the  figures  having  been 
made  upon  a  slate  or  portable  blackboard,  such  as  every  student  should 
have,  and  in  any  form  which  will  truly  represent  the  essential  opera- 
tions of  the  problems,  without  regard  to  the  particular  forms  of  the 
figures  in  the  book. 

The  recitations  may  properly  consist  of  mingled  interrogation  upon 
the  principles,  and  explanations  of  the  problems ;  the  latter,  both  from 
the  figures  in  the  book,  and  from  blackboard  constructions ;  also  from 
the  finished  constructions  on  the  student's  plates,  when  time  enough  is 
devoted  in  the  class-room  to  their  construction,  to  make  due  interroga- 
tion upon  them  possible.  It  will,  however,  probably  appear  on  trial 
that  no  test  equals  the  regular  class  recitation,  in  that  thoroughness 
which  is  due  to  the  student  in  his  training. 

It  is  a  peculiar  pleasure  to  add,  that  this  expensive  volume  now 
appears,  through  the  every  way  timely  and  pleasant  kindness  of  Students 
of  the  Institute,  to  the  Author,  and,  as  he  truly  hopes,  no  less  to  them- 
selves, in  making  up  a  liberal  subscription  in  aid  of  its  hitherto  delayed 
publication. 


7 


• 

SHADES  AND  SHADOWS. 


Book  1. 

CONSTEUCTION  OF  SHADES  AND  SHADOWS. 

PART  I. 
IN  COMMON  PROJECTION. 

SERIES  I. 

SHADES  AND  SHADOWS  FORMED  BY  PARALLEL  RATS. 

GENERAL  PRINCIPLES. 
§  1. — Definitions,  and  Classification  of  Problems. 

1.  Light  proceeds  from  its  source  in  uninterrupted  straight 
lines,  called  rays,  except  when  diverted  by  reflection  or  refrac- 
tion, or  arrested  by  opaque  bodies  ;  as  is  sufficiently  proved  by 
the  impossibility  of  direct  vision  through  a  bent  tube. 

2.  If  the  source  of  light  be  a  point,  the  volume  of  rays  pro- 
ceeding from  it,  and  intercepted  by  an  opaque  body,  will  be  a 
cone,  having  the  luminous  point  for  its  vertex.     If  this  point  be 
at  an  infinite  distance  from  the  opaque  body,  the  cone  becomes  a 
cylinder,  and  the  rays  of  light  will  be  sensibly  parallel.     The  latter 
case  is  the  one  considered  in  the  present  series  of  problems,  and 
is  substantially  the  same  as  that  of  bodies  illuminated  by  the 
Sun,  for,  on  account  of  his  great  distance  from  the  Earth,  the 
solar  rays  to  all  points  of  any  terrestrial  object,  are  sensibly  parallel. 

3.  Let  there  now  be  given,  in  the  following  order  of  successive 
position,  a  'Source  of  light,  an  opaque  point,  and  an  opaque  sur- 
face.    The  opaque  point  will  intercept  one  of  the  rays  from  the 
source  of  light     The  path  of  the  ray,  beyond  the  opaque  point, 

1 


2  GENERAL    PROBLEMS. 

will  therefore  be  a  line  of  darkness,  and  its  intersection  with  the 
opaque  surface  will  be  a  dark  point. 

The  line  of  darkness  is  the  shadow  in  space  of  the  given  point ; 
the  dark  point  on  the  opaque  surface  is  the  superficial  shadow, 
or,  simply,  the  shadow  of  the  same  point. 

4.  The  shadow  of  a  point  upon  any  surface  is,  therefore,  tho . 
intersection  of  a  ray  (meaning  the  indefinite  path  of  the  ray) 
drawn  through  the  point,  with  the  given  surface. 

5.  Any  plane  containing  a  ray,  is  a  plane  of  rays ;  for  any 
number  of  rays  can  be  drawn  in  the  plane,  and  parallel  to  the 
given  ray,  and  these  rays  will,  collectively,  constitute  a  plane  of 
rays.     Hence  a  plane  of  rays  can  be  drawn  through  a  straight 
line  having  any  position  in  space.     For  the  plane  containing 
this  line,  and  a  ray  through  any  point  of  it,  will  be  the  required 
plane  of  rays.     Furthermore,  since  two  lines  determine  a  plane, 
but  one  plane  of  rays  can  be  passed  through  a  given  line,  unless 
the  line  be  parallel  to  the  rays  of  light. 

6.  A  line  being  made  up  of  points,  its  shadow  will  consist  of  the 
shadows  of  all  its  points.     Eays  of  light  being  parallel,  the  rays 
through  all  the  points  of  a  straight  line,  will  form  a.  plane  of  rays 
(5),  unless  the  line  coincides  with  a  ray.     The  shadow  of  a  straight 
line  is,  therefore,  in  general,  the  line  of  intersection  of  a  plane 
of  rays  passed  through  it,  with  the  surface  receiving  the  shadow. 
When  the  line  coincides  with  a  ray,  its  shadow  is  its  own  point 
of  intersection  with  the  given  surface. 

When  a  line  is  parallel  to  any  plane,  its  shadow  on  that  plane 
is  parallel  to  the  line  itself. 

The  shadows  of  parallel  lines,  on  the  same  plane,  are  parallel. 
Also  the  shadows  of  the  same  line  on  parallel  planes  are  parallel ; 
only  one  of  them,  however,  can  be  real. 

7.  The  parallel  rays  through  all  points  of  a  curve,  will  form  a 
cylinder  of  rays,  unless  the  curve  be  plane,  and  coincides  with  a 
plane  of  rays.     Hence  the  shadow  of  a  curved  line  is,  in  general, 
the  intersection  of  the  cylinder  of  rays  having  the  curve  for  its 
directrix,  with  the  given  opaque  surface.     When  the  curve  is 
plane,  and  in  a  plane  of  rays,  its  shadow  is  the  line  in  which 
that  plane  intersects  the  opaque  surface. 

8.  The  shadow  of  a  surface  is  a  surface,  unless  it  is  a  plane 
surface  and  coincides  with  a  plane  of  rays,  when  its  shadow  will 
be  a  line,  as  in  the  case  of  a  plane  curve  lying  in  a  plane 
of  rays. 


SHADES  AND  SHADOWS.  3 

9.  The  shadow  of  a  solid  is  always  a  surface,  which  is  known 
when  its  boundary,  called  its  line  of  shadow,  is  found.     Let  the 
given  body  be  circumscribed  by  parallel  tangent  rays.     These 
rays  will  form  a  circumscribing  tangent  cylinder  of  rays.     The 
curve  of  contact  of  this  cylinder  and  the  body,  is  evidently  the 
boundary  between  the  side  of  the  body  towards  the  source  of 
light,  and  the  part  which  is  opposed  to  the  light.     The  former  is 
the  illuminated  part  of  the  body,  the  latter  is  the  shade  of  the 
body,  and  the  line  of  contact  is  called  the  line  of  shade. 

10.  Therefore  1st.  From  all  the  portion  of  space  within  the  tan- 
gent cylinder  of  rays,  and  beyond  the  given  body,  light  is  exclud- 
ed, and  that  portion  is  called  the  shadow  in  space,  of  the  body. 
2nd.  The  area,  bounded  by  the  intersection  of  the  cylinder  of 
rays  with  the  surface  receiving  the  shadow,  is  the  shadow  of  the 
body  on  that  surface.     3d.  The  line  of  shadow  is  thus  the  shadow 
of  the  line  of  shade.     4th.  The  line  of  shade  on  a  body  must 
therefore  be  known,  before  its  line  of  shadow  can  be  found. 

11.  Any  one  point  of  a  curve  of  shade  upon  a  body,  is  the 
point  of  contact  of  one  element  of  the  tangent  cylinder  of  rays ; 
that  is  of  one  ray :  and  the  intersection  of  tliat  ray,  with  any 
surface  beyond  the  given  body,  is  the  shadow  of  that  one  point  of 
the  curve  of  shade.     Both  shades  and  shadows  are  found,  in 
practice,  according  to  this  statement,  i.e.  by  separately  finding 
single  points  and  joining  them. 

12.  The  constitution,  and  mode  of  construction,  of  the  line  of 
shade  of  a  body,  will  depend  on  the  nature  of  the  surface  of 
that  body.     The  line  of  shade  on  a  plane  sided  body,  is  com- 
posed of  those  edges,  collectively,  which  divide  its  light,  from  its 
shaded  surfaces.     On  a  developable  single  curved  surface,  it  con- 
sists of  the  elements  of  contact  of  tangent  planes  of  rays.     On 
a  warped  surface,  it  consists  of  a  succession  of  points,  one  on 
each  of  a  succession  of  elements,  these  points  being  the  points 
of  tangency  of  planes  of  rays.     On  a  double  curved  surface,  it 
consists  strictly  of  the  curve  of  contact  of  a  circumscribing  tan- 
gent cylinder  of  rays. 

The  problems  of  shades  may  therefore  be  arranged  in  the  order 
of  the  four  kinds  of  surfaces  as  just  named. 

13.  Every  problem  of  shadows,  evidently  reduces  to  finding 
the  shadow  of  a  point  on  a  given  surface  ;  that  is,  to  finding  the 
intersection   of  a  straight  line   (ray)   with  the  surface.     The 
method  of  finding  this  intersection  depends  on  the  nature  of  the 


4:  GENERAL    PROBLEMS. 

given  surface,  hence  the  problems  of  shadows  may  be  arranged  in 
the  same  way  as  those  of  shades  may  be. 

But  the  construction  of  the  shadows  of  straight  lines  being 
generally  a  little  simpler  than  that  of  the  shadows  of  curves, 
the  surface  receiving  the  shadow  being  the  same  in  both  cases, 
the  former  may  be  made  introductory  to  the  latter,  in  each  of  the 
four  principal  groups  of  problems  of  shadows. 

14.  Rehearsing  the  principal  conclusions  thus  far  reached,  we 
have  the  following : 

The  curve  of  shade  of  a  body  is  the  curve  of  contact  of  a  tangent 
cylinder  of  rays,  with  that  body.  The  curve  of  shadow  of  the 
same  body,  is  the  curve  of  intersection  of  the  same  cylinder  with 
the  surface  receiving  the  shadow. 

Any  one  point  of  the  curve  of  shade,  is  the  point  of  contact  of  one 
element  of  the  cylinder,  and  this  element  is  a  ray  of  light.  The 
intersection  of  the  same  ray,  or  element,  with  the  surface  receiving 
the  shadow,  is  the  shadow  of  this  same  point  in  the  curve  of  shade. 

15.  Here,  expanding   the  last  article  somewhat,  we  observe 
that  the  line  of  shade  is  either  straight  or  curved,  hence  the  sur- 
face of  rays  passed  through  it — and  whose  intersection  with  any 
other  surface  is  the  line  of  shadow — will  always  be  a  plane  when 
the  line  of  shade  is  straight,  and  single  curved  when  the  line  of 
shade  is  a  curve.     In  the  latter  case  the  line  of  shade  will  be  the 
directrix  of  the  single  curved  surface  of  rays.     When  the  rays 
are  parallel,  the  single  curved  surface  of  rays  will  be  cylindrical ; 
•when  they  are  diverging,  it  will  be  conical,  having  the  luminous 
point  for  its  vertex.     Hence,  for  all  combinations  of  straight  and 
curved  lines  of  shade,  with  parallel  rays,  and  rays  diverging 
from  a  point,  the  curve  of  shadow — defined  as  found  by  the 
problems  of  orthographic  projection — will  be  the  intersection  of  a 
plane,  a  cylinder,  or  a  cone,  with  the  surface  receiving  the  shadow. 

16.  The  case,  in  which  the  source  of  light  is  a  luminous  poin t, 
either  at  a  finite  or  infinite  distance,  is  the  only  one  which  needs 
to  be  considered.     For,  by  way  of  a  simple  example  of  simul- 
taneous illumination  from  several  points,  let  a  sphere  be  exposed 
to  the  light  from  a  finite  luminous  straight  line.     Each  point  of 
such  a  line  would  be  the  vertex  of  a  separate  tangent  cone  of 
rays,  whose  circle  of  contact  with  the  sphere  would  be  the  circle 
of  shade  due  to  that  cone.     Hence,  only  so  much  of  the  sphere 
os  received  no  light  from  either  extremity  of  the  luminous  line, 
would  be  totally  in  shade ;  and  only  so  much  as  received  light 


SHADES  AND   SHADOWS.  5 

from  every  point  of  the  line  would  be  totally  illuminated.  The 
intermediate  portion  would  be  in  partial  shade,  which  would 
become  more  and  more  complete  as  the  total  shade  should  be 
approached.  In  order,  therefore,  to  secure  both  simplicity  and 
precision  in  the  diagrams  of  shades  and  shadows,  the  source  of 
light  is  assumed  to  be  a  point. 


§  2. —  Graphical  Representation  of  Shades  and  Shadows. 

17.  In  passing  to  the  graphical  construction  of  shadows,  it  is 
to  be  observed,  that,  in  representing  them  in  orthographic  projec- 
tion, the  planes  of  projection,  the  projecting  lines,  and  the  use  of 
these,  are  similar  to  the  same  things,  and  their  use,  as  found  in 
any  other  problems  of  orthographic  projection. 

18.  The  three  things  given  in  space  (3)  for  the  production  of  a 
real  shadow,  are  given  in  projection,  in  representing  that  shadow ; 
only,  as  the  source  of  light  is  supposed  to  be  indefinitely  remote, 
it  is  only  virtually  represented,  by  the  projections  of  the  parallel 
rays  proceeding  from  it. 

If  the  source  of  light  were  a  near  luminous  point,  its  projec- 
tions would  be  given,  together  with  rays  proceeding  from  it  in 
diverging  lines. 

19.  The  ordinary  rules  of  orthographic  projection,  concerning 
given  and  auxiliary,  hidden  and  visible  lines  and  planes,  are 
observed  in  problems  of  shades  and  shadows.     Moreover,  visible 
shades  or  shadows  are  represented,  either  by  a  tint  of  indian  ink 
or  by  parallel  lines.     Hidden  shades  or  shadows  are  represented 
only  by  their  boundaries,  which  are  made  in  dotted  lines ;  plain, 
or  fringed,  as  in  PI.  L,  Fig.  5,  for  greater  distinctness. 

20.  In  constructing  shadows  merely  as  a  geometrical  exercise, 
the  main  object  is  to  indicate  clearly  their  position.     Neatness 
and  distinctness  are,  therefore,  all  that  is  required  in  their  exe- 
cution ;  hence,  to  save  time,  they  are  represented  in  flat  or  uni- 
form tints.    In  making  finished  drawings,  the  shades  and  shadows 
are  done  in  graduated  tints, according  to  their  actual  appearance. 

21.  In  industrial  applications  of  problems  of  shades  and  sha- 
dows, the  light  is  usually  taken  to  correspond  with  the  body 
diagonal  of  a  cube,  two  of  whose  faces  coincide  with  the  planes 
of  projection.     That  is,  the  light  coming  from  the  left,  forward 
and  downward,  makes  an  angle  of  35°  16'  with  the  planes  of 


6  GENERAL  PROBLEMS. 

projection,  and  its  projections  make  angles  of  45°  with  the 
ground  line. 

In  the  following  general  problems,  the  light  is  taken  indif- 
ferently in  any  direction,  since  the  general  methods  of  solution 
are  not  altered  by  so  doing. 

22.  Methods  of  solution  are  distinguished  as  General  and  /Spe- 
cial, also  as  Direct  and  Indirect. 

General  methods  consist  of  such  operations  as  apply  equally 
well,  in  theory,  though  not  always  with  equal  practical  conve- 
nience, to  whole  classes  of  problems. 

^Special  methods  are  such  as  are  founded  on  the  distinguishing 
peculiarities  of  individual  problems,  or  groups  of  problems.  They 
are  therefore  as  numerous  as  those  peculiarities,  and  hence  afford 
the  natural  field  for  the  exercise  of  ingenuity  in  their  discovery 
and  application. 

Direct  methods  are  those  which  immediately  and  obviously 
conform  to  the  usual  definition  of  a  shade  or  a  shadow  ;  as  when 
a  point  of  shade  is  found  as  the  point  of  contact  of  a  given  ray 
with  a  given  surface. 

Indirect  methods  are  those  in  which  auxiliary  magnitudes  are 
employed,  to  avoid  the  laborious  constructions  which  direct 
methods  may  sometimes  occasion ;  as  when  we  find  a  point  of 
shadow  on  some  irregular  surface,  by  noting  where  the  easily 
found  shadow  upon  some  secant  plane  meets  the  intersection  of 
that  plane  with  the  surface. 


SHADES  AND   SHADOWS. 


SERIES  I. 

SHADES    AND    SHADOWS    FORMED    BY   PARALLEL    RAYS 

DIVISION  I. 
SHADES  AND  SHADOWS  ON  RULED  SURFACES. 

23.  The  line  of  shade  on  any  ruled  surface  is  obtained  by  find- 
ing, either  by  inspection,  or  by  construction — by  means  of  tangent 
rays  or  planes  of  rays — the  edges,  elements,  or  points  which 
constitute  that  line  of  shade. 

24.  A  shadow  cast  by  a  body  upon  any  ruled  surface,  is  deter- 
mined by  finding  an  element  of  that  surface,  and  a  point  of  the  line 
of  shade  of  the  body,  both  of  which  shall  be  in  the  same  plane 
of  rays.     A  ray  through  the  point  of  shade  will  then  intersect 
the  element  of  the  ruled  surface  in  a  point  of  the  required  shadow. 

CLASS  I. 

SHADES  AND  SHADOWS  ON  PLANE  SURFACES. 
§  I. — Shades. 

25.  The  shade  of  an  opaque  plane,  or  plane  figure,  is  the  whole 
of  one  side  of  such  plane  or  plane  figure.     The  former  being  of 
indefinite  extent,  can  have  no  line  of  shade.     That  of  the  latter 
is  its  perimeter. 

26.  The  line  of  shade  of  any  plane-sided  body,  as  a  pyramid, 
consists  of  those  edges,  collectively,  which  divide  its  illuminated 
from  its  unilluminated  surfaces  (12).     This  line  of  shade  can 
often  be  determined   by  inspection,  having  given  the  relative 
position  of  the  body  and  the  rays  of  light. 

The  two  following  problems  will  show  the  method  of  finding 
lines  of  shade  on  plane-sided  bodies,  when  they  cannot  be  deter- 
mined by  inspection. 


8  GENERAL   PROBLEMS. 


PROBLEM  I. 

To  determine,  first,  ivhether  the  line  of  intersection  of  two  oblique 
planes  is,  or  is  not,  a  line  of  shade  ;  and,  second,  to  determine,  by 
inspection,  the  compound  line  of  shade  on  a  given  plane-sided  body. 

First: — Principles. — After  finding  the  projections  of  the  line  of 
intersection  of  the  given  planes,  by  Prob.  V.,  Des.  Geom.,  assume 
any  point  on  that  line,  and  draw  the  projections  of  a  ray  of  light 
through  it.  If  the  ray,  thus  given,  pierces  either  plane  of  pro- 
jection between  the  traces  of  the  given  planes  on  that  plane,  it 
shows  that  both  planes  are  illuminated,  both  being  pierced  by 
rays,  and  hence  that  their  intersection  is  not  a  line  of  shade.  On 
the  contrary,  if  the  ray  pierces  either  plane  of  projection  outside 
of  the  traces  of  the  given  planes  on  that  plane,  it  shows  that  the 
intersection  of  the  given  planes  is  a  line  of  shade. 

Construction. — PI.  I.,  Fig.  1.  Let  RS  be  the  ground  line, 
PRP'  one  of  the  given  planes,  PSP'  the  other,  and  Pf—P'e' 
their  line  of  intersection,  n,  n'  is  an  assumed  point  of  this  inter- 
section, through  which  the  ray,  na—n'a',  parallel  to  the  given 
ray,  r—r',  is  drawn.  This  ray  pierces  the  horizontal  plane  at 
a,  a',  indicating,  according  to  the  principles  of  this  solution,  that 
Pf—P'e'  is  a  line  of  shade. 

Remarks. — a.  Hence  the  projections  of  the  visible  portion  of 
the  unillumiuated  plane,  PSP',  are  shaded,  and  Pf—P'e'  is  inked 
as  a  heavy  line. 

b.  The  above  construction  is  applicable  to  the  case  of  any 
pyramid,  in  a  simple,  or  oblique  position,  when  we  wish  to  deter- 
mine its  line  of  shade.  In  such  cases,  unless  the  pyramid  rests 
upon  a  given  plane,  it  will  first  be  necessary  to  determine  the 
traces  of  its  faces  on  some  plane,  whose  intersections  with  the 
rays  through  points  in  the  edges  of  those  faces  will  also  be  con- 
structed. 

27.  As  a  second  illustration  of  this  problem,  see  PI.  IV.,  Fig. 
14,  where  PQP'  and  PRP'  are  the  given  oblique  planes.  PV 
and  Pb  are  the  projections  of  their  intersection,  and  m,m  is  a 
point  on  that  line,  mn—m'n'  is  a  ray  of  light  through  m,m', 
and  it  pierces  the  horizontal  plane  at  n,  outside  of  both  planes. 
Jlenc'-,  when  Ph  —  P'a'  is  regarded  as  the  edge  of  the  salient  die- 
ilral  angle,  indicated  by  the  portions,  QP  and  RP,  of  the  traces, 
it  is  an  edge  of  shade  for  the  direction,  mn—m'n'  of  the  light. 


SHADES  AND   SHADOWS.  9 

If  PQP'  as  now  shown,  and  that  part  only  of  PBP',  extending 
in  front  of  the  edge  P6  —  PV,  are  considered  as  the  surfaces 
exposed  to  the  light,  the  left  side  of  each  plane  is  then  the  one 
considered,  and  both  of  these  surfaces  are  in  the  light. 

The  next  problem  affords  a  more  practical  example  of  th 
general  constructions  just  given. 

28.  Second. — The  lines  of  shade  on  the  abacus  and  pillar. — See 
PI.  I.,  Fig.  4,  where  the  projections  of  a  square  pillar  and  its 
abacus,  resting  on  the  horizontal  plane,  are  given,  together  with 
the  direction  of  the  light,  Cm — D'ra'. 

Here  it  is  evident  that  t  —  t"t' ;  C  —  C'D';  the  upper  edges 
CE  and  E A ;  the  vertical  edges  at  A  and  q ;  AB  —  A/B',  and 
BC  —  B'C',  are  the  edges  which  constitute  the  compound  line  of 
shade  of  the  given  body.  They  are  therefore  (9 — 11)  the  edges 
which  cast  shadows  on  the  pillar,  or  on  other  surfaces  ;  and  they 
are  identical  with  the  lines,  which,  in  geometrical  line-drawings, 
are  inked  as  heavy  lines. 


PROBLEM  II. 

To  determine  the  edges  of  shade  on  a  pyramid,  whose  axis  is  oblique 
to  both  planes  of  projection. 

Construction. — PI.  IV.,  Fig.  15.  To  make  the  problem  quite 
definite,  let  Gr"L"  be  the  ground  line  of  an  auxiliary  vertical 
plane,  parallel  to  the  axis  of  the  pyramid,  which  we  will  suppose 
to  be  a  triangular  one,  in  order  to  avoid  needless  repetition  of  the 
same  operations  of  construction.  Let  the  plane  of  the  base  be 
perpendicular  to  the  axis,  and  let  a"c"  be  its  trace  on  the  auxi- 
liary vertical  plane.  Let  ABC  be  the  true  size  and  form  of  this 
base,  seen  when  its  plane  is  revolved  about  a"c",  as  its  trace,  into 
the  auxiliary  vertical  plane,  and  let  D  be  the  projection  of  the 
axis  on  the  plane  of  the  base.  Then  by  counter-revolution,  A,B 
and  C  return  in  arcs,  projected  in  Aa",  etc.,  perpendicular  to 
a"c",  to  the  points  a",  b"  and  c  ' ;  and  D,  the  foot  of  the  axis,  to 
d".  Then  make  d"v",  perpendicular  to  a"c",  equal  to  the  axis 
of  the  pyramid,  and  join  v"  with  a",  b"  and  c"  to  complete  the 
auxiliary  projection. 

To  make  the  horizontal  projection,  make  aa",  etc.,  equal  to 
Aa",  etc.,  and  perpendicular  to  the  ground  line  Gr"L"  ;  and  v  — 


10  GENERAL  PROBLEMS. 

abc  will  be  the  horizontal  projection  of  the  pyramid.  The  vertical 
projection,  v'  —  a'b'c',  is  found,  as  in  all  similar  cases,  by  pro- 
jecting up  v  —  b,  etc.,  in  perpendiculars  to  GL,  and  by  making 
the  heights  of  the  points  v',  —  &',  etc.,  above  the  ground  line  GL, 
equal  to  their  heights  v"k,  b"h,  etc.  above  G"L". 

Now,  to  test  any  edge  of  the  pyramid,  to  see  if  it  be  an  edge 
of  shade,  or  not. 

First  Method. — Let  the  edge  vb  —  v'b'  be  taken.  Find  the 
traces,  on  one  plane  of  projection,  of  the  planes  of  the  faces  of 
which  the  given  edge  is  the  intersection.  Then  draw  a  ray 
through  any  point  of  that  edge,  and  if  it  meets  the  plane  of  the 
traces  outside  of  those  traces,  when  the  edge  belongs  to  a  salient 
angle  turned  towards  the  light,  the  given  edge  is  a  line  of  shade. 
Accordingly,  vb  —  v'b'  pierces  the  horizontal  plane  at  m,  and 
cb  —  c'b'  pierces  it  at  n,  giving  mn,  as  the  horizontal  trace  of  the 
plane  of  the  face  vbc  —  v'b'c'.  Again,  va  —  v'a'  pierces  the  hori- 
zontal plane  at  a,  giving  ma  as  the  horizontal  trace  of  the  plane 
of  the  face  vab  —  v'a'b'.  Now  a  ray,  bp  —  b'p',  through  any  point 
bb'  of  the  intersection,  vm  —  v'm',  of  these  planes,  pierces  the 
horizontal  plane  atjp,  outside  of  both  traces,  ma  and  inn.  But 
the  parts  of  these  planes  which  are  the  faces  of  the  pyramid,  esti- 
mated from  the  common  edge,  vb  —  v'b',  have  ma  and  mO  for 
their  horizontal  traces.  Hence  the  direction  of  the  light  evidently 
strikes  only  the  interior  face  of  each  of  these  planes,  that  is,  the 
face  towards  the  axis  of  the  pyramid ;  so  that  the  exterior  sur- 
faces vbc  and  vba  are  both  in  the  dark,  and  vb  —  v'b'  is  not  a  line 
of  shade. 

If  now  we  find  0,  the  intersection  of  the  edge,  vc  —  v'c'  with 
the  horizontal  plane,  aO  is  the  horizontal  trace  of  the  plane  of 
the  face  vca  —  v'c'a',  and  vcO  is  a  salient  edge,  similar  to  Pb  — 
PV  in  Fig.  14,  so  that  the  ray,  cr  —  cV,  by  piercing  the  hori- 
zontal plane  outside  the  traces  On  and  Oa,  shows  that  vc  —  v'c' 
is  an  edge  of  shade,  which,  indeed,  was  known  as  soon  as  it  was 
found  that  vb  —  v'b'  was  not  such  an  edge. 

Second  Method. — Find  the  shadows  of  all  the  edges  of  the 
pyramid  on  either  plane  of  projection.  Then,  if,  for  example, 
the  shadow  of  vc  —  v'c'  upon  the  horizontal  plane  is  further  from 
the  pyramid,  in  the  direction  of  the  light,  than  that  of  vb  —  v'b', 
it  would  show  that  vc  —  v'c',  only,  of  those  two  edges,  really 
cast  a  shadow.  Hence  it  would  be  shown  that  vc  —  v'c'  was  an 
edge  of  shade  (26).  That  is,  we  have  this  principle:  TJiosc 


SHADES  AND  SHADOWS.  11 

edges,  whose  shadows  are  the  boundaries  of  the  real  shadow  of  a  body, 
are  the  edges  of  shade  of  that  body. 

Bemark. — The  next  problem,  if  not  the  preliminary  general 
principles  already  given,  will  enable  the  student  to  construct  this 
second  method. 

§  H.  Shadows. 

29.  Here  observe — First :  That  whenever  a  plane  is  perpen- 
dicular to  either  plane  of  projection,  its  entire  surface,  and  all 
contained  in  it,  is  projected  in  the  trace  of  the  plane  on  that 
plane  of  projection.     /Second:  Having  a  point,  p,  which  casts  a 
shadow  on  a  plane,  the  projections  of  that  shadow  must  be  in 
the  projections  of  the  ray  (4)  through  p. 

30.  From  the  two  foregoing  principles  we  have  the  following 
simple  general  method  (22)  of  finding  points  of  shadow  on  the 
planes  of  projection,  or  on  any  planes  which  are  perpendicular  to 
them.     Find,  by  inspection,  the  intersection  of  the  linear  projection 
of  the  plane  receiving  the  shadow,  ivith  the  corresponding  projection 
of  the  ray  through  any  point  casting  a  shadow  upon  it.     This  will 
be  one  projection  of  the  point  of  shadow  sought.     Its  other  projection 
will  be  the  intersection  of  the  projecting  line  of  the  point  found,  with 
the  other  projection  of  the  same  ray. 

It  is  only  necessary  here  to  remember,  that  the  linear  projection 
of  both  planes  of  projection,  is  the  ground  line. 


PROBLEM  III. 

To  find  the  shadow  of  a  straight  line,  on  the  horizontal  plane  of 

projection. 

Principles. — The  shadow  of  a  straight  line  on  a  plane,  is  the 
intersection  of  that  plane  with  a  plane  of  rays  passed  'through 
the  given  line  (5-6).  This  shadow  will  therefore  be  a  straight 
line.  But  two  points  determine  a  straight  line,  and  two  parallel 
lines  determine  a  plane.  Hence  if,  through  any  two  points  of 
the  given  line,  we  pass  rays,  their  intersection  with  the  given 
plane  will  determine  the  shadow  of  the  given  line.  Also,  the 
point  in  which  the  given  line  meets  the  given  plane  is  a  point  of 
the  required  shadow,  for  the  line  and  its  shadow  are  in  the  same 
plane  (of  rays)  and  therefore  intersect. 


12  GENERAL  PROBLEMS. 

Construction. — PI.  L,  Fig.  2.  Let  ab  —  a'b'  be  the  given  line, 
and  r —  r  the  direction  of  the  light.  By  (30)  the  ray  ac  —  a'c', 
drawn  through  any  assumed  point,  as  a,  a'  of  the  given  line, 
pierces  the  horizontal  plane  in  the  point  whose  vertical  pro- 
jection is  c',  and  whose  horizontal  projection  is  therefore  c. 
Likewise  the  ray  bd — b'd'  pierces  the  horizontal  plane  at  d',d 
(naming  that  projection  first  which  is  found  first).  Hence  cd  is 
the  shadow  of  ab  —  a'b'  on  the  horizontal  plane.  Also  e,e',  the 
point  in  which  ab  —  a'b'  pierces  the  horizontal  plane,  is  a  point 
of  its  shadow  on  that  plane,  and  hence  will  be  found  in  dc  pro- 
duced. 

Remark. — In  taking  a  strict  view  of  the  principles  of  this  solu- 
tion, and  in  similar  cases,  it  should  be  observed  that  the  rays,  as 
ah— a'b',  are  to  be  regarded  as  cut  from  the  plane  of  rays  (6) 
passed  through  the  line,  by  secant  planes  of  rays,  which  may 
have  any  position,  but  are  most  readily  conceived  of  as  perpen- 
dicular to  one  of  the  planes  of  projection. 

31.  When  a  line  is  perpendicular  to  a  plane,  its  shadow,  on  that 
plane,  is  in  the  direction  of  the  projection  of  the  light  on  that  plane  j 
for  the  plane  containing  the  line  and  a  ray  through  any  point  of 
it,  is  a  plane  of  rays  (5),  perpendicular  to  the  given  plane,  and  is 
therefore  a  projecting  plane  of  all  rays  intersecting  the  line. 
Hence  its  trace  on  the  given  plane,  is  both  the  shadow  of  the  line 
(6),  and  the  projection  of  a  ray,  upon  that  plane. 

This  principle  is  of  very  frequent  application  to  the  shadows 
of  vertical  lines  upon  the  horizontal  plane,  and  of  co- vertical  lines 
(that  is,  of  lines  perpendicular  to  the  vertical  plane)  upon  the 
vertical  plane. 

PROBLEM  IV. 

To  construct  the  shadow  of  a  square  abacus  upon  a  square  pillar, 
and  of  both,  on  the  horizontal  plane  of  projection. 

[There  being  no  new  principles  in  this  problem,  we  pass  at 
once  to  its  graphical  solution  as  a  new  illustration  of  the  method 
of  (30)  referring  to  (28)  for  the  line  of  shade.] 

Construction. — PI.  I.,  Fig.  4.  Let  the  body  be  seen  obliquely 
as  shown  on  the  vertical  plane,  and  let  Cm—D'm'  be  the  direc- 
tion of  the  light.  The  point  d,d',  found  by  drawing  the  ray  qd, 
and  projecting  d  at  d',  casts  the  point  of  shadow  q,q'  on  the  ver* 


SHADES  AND   SHADOWS.  13 

tical  edge  at  q ;  q'  being  found  at  the  intersection  of  this  edge 
with  the  vertical  projection,  d'q',  of  the  raj  through  d,d'.  From. 
q'  the  shadow  q'c'  is  parallel  to  the  line  de—d'e',  which  casts  it, 
since  that  line  is  parallel  to  the  vertical  face,  qc,  of  the  pillar  (6). 
The  ray  through  B,B'  pierces  the  front  of  the  pillar  at  </,  whose 
vertical  projection  is  in  the  vertical  projection,  By,  of  the  ray,  at 
gr.  Hence  c'g'  is  the  shadow  of  the  portion,  eB  —  e'B',  of  the 
edge,  AB— A'B',  of  the  abacus,  upon  the  front  of  the  pillar. 
g'f,  parallel  to  B'C',  is  the  shadow  of  the  portion  B/-B/  of  the 
edge  BC— B'C',  on  the  parallel  front  of  the  pillar.  The  ray 
through  //',  passing  beyond  the  pillar,  pierces  the  horizontal 
plane  (30)  at  hf,  A,  and  hi  is  the  shadow  of  the  portion,  t—t'tf',  of 
the  edge  of  the  pillar,  which  casts  a  shadow  on  the  horizontal 
plane.  The  portion  JV—f'W  of  BC-  B'C'  casts  the  shadow  hk, 
beginning  at  h  and  limited  at  k',k  by  the  ray  Ck—C'k.  The 
edge  C— C'D'  casts  the  shadow  km,  limited  by  the  intersection  of 
the  rays  Ck— C'k'  ancUC/n— D'm'  with  the  horizontal  plane. 
Similarly,  CE  has  rar?,  equal  and  parallel  to  itself,  for  its  shadow ; 
E  A  has,  likewise,  no  for  its  shadow  ;  the  vertical  edge  at  A  deter- 
mines the  shadow  op,  found  as  km  was ;  pr  is  the  shadow  of 
Ad—  AW,  and  rq  is  the  shadow  of  the  vertical  edge  of  the  pillar 
at  q. 

PROBLEM  Y. 

Having  the  projections  of  a  ray  of  light,  upon  two  vertical  planes  at 
right  angles  to  each  other  ;  to  find  its  projections  on  those  planes, 
after  revolving  it  90°  about  a  vertical  axis. 

Principles. — When  we  view  any  object  in  reference  to  a 
given  vertical  plane,  the  direction  of  the  light  is  regarded  as 
fixed ;  and  it  is  therefore  fixed  in  projection,  while  we  are  find- 
ing the  shadow  of  the  object  on  that  plane.  When,  however, 
we  turn  and  face  the  object  in  a  different  direction,  the  light 
may  be  supposed  either  to  turn  with  us,  so  as  to  come  in  the 
same  relative  direction  with  respect  to  the  new  projection  that  it 
did  with  respect  to  the  old;  or  it  may  be  supposed  to  remain 
absolutely  fixed,  as  it  is  practically  in  Nature  for  any  short 
interval.  A  little  reflection  on  the  latter  supposition,  will  show 
that,  in  certain  aspects  of  an  object,  only  its  shade  could  be  seen  ; 
hence  it  is  sometimes,  if  not  generally,  desirable  to  adopt  the  first 
view,  viz.  that  the  light  turns  with  us,  so  as  to  come  in  the  same 


14:  GENERAL  PROBLEMS. 

relative  direction  with  respect  to  each  new  vertical  plane  of  pro* 
jection.  This  first  view  is  adopted  in  the  following — 

Construction. — See  PI.  II.,  Fig.  6.  Here  let  the  given  projec- 
tions of  the  light,  on  the  two  vertical  planes,  be  R'L'  and  R"L", 
and  let  qpx'  indicate  the  real  position  of  the  vertical  plane  of  the 
left  hand  projection.  By  an  obvious  construction,  we  find  RL, 
the  horizontal  projection  of  the  given  ray,  R^L"  — RT/.  Then, 
as  the  observer  turns  90°  to  face  the  plane  qp,  the  light  turns 
with  him,  an  equal  amount,  about  any  vertical  axis,  as  at  L,  and 
so  appears  in  horizontal  projection,  as  at  rL,  perpendicular  to  RL. 
The  vertical  projections  of  rL  are,  obviously,  r'l/  and  r"~L". 

We  see  now,  by  inspection,  that,  as  the  vertical  planes,  and 
also  the  two  positions  of  the  light,  as  seen  at  RL  and  rL,  are  at 
right  angles  to  each  other,  pt=sm,  and  pq=mn.  That  is,  x'y' 
being  the  same  for  all  the  vertical  projections  of  the  ray,  the  new 
vertical  projections,  r'L'  and  r"L",  make  the  same  angle  with  the 
ground  line  as  R'L'  and  R"L"  do.  So  that,  in  practice,  it  is 
only  necessary  to  draw  rT/  parallel  to  R"L",  and  r"L"  equally 
inclined  with  R'L', — but  in  an  opposite  direction — in  order  to 
obtain  the  projections  of  the  light  as  it  proceeds  in  viewing  the 
side  or  left  hand  elevation. 

When  the  projections,  r?— rT,  of  the  light  make  the  conven- 
tional angle  of  45°  with  the  ground  line,  as  in  PI.  II.,  Fig.  7, 
the  vertical  projection,  r""l"",  of  it  on  the  right  hand  plane,  as 
it  comes  when  the  observer  has  turned,  from  viewing  that  plane, 
to  view  the  left  hand  one,  simply  inclines  at  45°  to  the  ground 
line  in  the  opposite  sense,  from  r'l'. 

PROBLEM  VI. 

To  find  the  projections  of  the  shadows  on  the  parts  of  a  timber  fram- 
ing, shown  in  two  elevations,  on  two  vertical  planes  at  right  angles 
to  each  other. 

Principles. — The  chief  new  principle  in  this  problem  is,  that  a 
point  of  shadow  may  exist,  geometrically,  on  a  material  surface 
produced ;  though,  physically,  only  on  the  real  portion  of  that 
surface. 

Construction.— PI.  II.,  Fig.  8.  MJ-J'J"  is  a  horizontal 
timber,  resting  on  the  transverse  timber  AJ— A'B'C',  and 
on  the  lateral  braces  kh—g'y  and  kJt—nu.  These  braces  are 


SHADES  AND  SHADOWS.  15 

framed  into  the  post  LH— A'H',  in  front  of  whicja  is  the  front 
brace  GP — GT'.  The  left  hand  elevation  is  supposed  to  be 
made  on  a  vertical  plane  at  right  angles  to  the  plane  of  the  paper. 
Taking  the  light  in  this  practical  problem  in  the  conventional  di- 
rection (21)  GL— G'l/  are  the  projections  of  a  ray,  as  it  proceeds 
in  viewing  the  object  as  seen  on  the  right  hand  plane  of  projection. 

This  being  established,  we  are  prepared  to  construct  the  sha- 
dows in  Fig.  8.  The  point  A,  A' casts  a  shadow  on  the  front 
brace  at  a,  whose  other  projection,  a',  is  the  intersection  of  the 
projecting  line,  a— a',  with  the  other  projection,  AV,  of  the  ray 
Aa.  A  —  A'B',  being  parallel  to  the  face  of  this  brace,  its  shadow, 
a' a",  is  parallel  to  A— A'B'.  Next,  c,c'  and  L,L/,  the  shadows 
of  the  points  a,a"  and  G,G'  in  the  front  right  hand  edge  of  the 
brace,  are  found  precisely  as  a,a'  was,  and  they  determine  LY,  the 
position  of  the  imaginary  shadow  (in  this  figure)  of  this  edge  on 
the  plane  of  the  front  of  the  post  produced.  The  shadow  of  the 
back  left  hand  edge,-  which  pierces  the  post  at  P,P',  begins, 
therefore,  at  P,P'  and  is  parallel  to  LY,  as  seen  at  P'K.  The 
front  of  the  lateral  brace,  g'y,  being  parallel  to  the  front  of  the 
post,  the  shadow  of  Ga  —  G'a"  upon  it,  will  be  parallel  to  LV. 
One  point  of  this  shadow  is  b,b",  found  by  drawing  the  ray  ab — 
a"b"  ;  hence  b"f,  parallel  to  LY,  is  the  required  shadow.  The 
point  A,  B'  casts  the  shadow  5,6',  found  like  the  preceding,  on 
the  plane  of  the  front  of  the  lateral  brace.  Then  6Y,  parallel  to 
B'C',  and  limited  by  the  ray  through  C,C',  is  the  shadow  of  AC — 
B'C'  on  the  same  plane.  The  portion  of  shadow  on  the  brace, 
bounded  by  CY,  is  cast  by  the  edge,  CJ — C',  of  the  transverse 
timber,  and  b'b"  is  the  shadow  of  A — A'B'  on  the  plane  of  the 
front  of  this  brace.  Finally,  for  the  right  hand  elevation,  the  edge 
J — 3' 5"  casts  the  shadow  n — y  on  the  lateral  braces,  found  by 
projecting  over  k",  where  a  ray  through  any  point  of  this  edge 
meets  the  plane  of  the  front  of  these  braces. 

Proceeding  to  the  side  elevation,  two  planes  of  rays,  containing 
the  edges  whose  projections  are  J  and  M,  will  include  between 
them  the  shadow  in  space  (10)  of  the  longitudinal  timber  JM — 
J'J".  Their  traces  on  the  side  of  the  transverse  timber,  will 
therefore  bound  the  shadow  on  that  timber.  A  portion  of  one 
of  these  traces  is  seen  at  Jk".  The  edge  AB — B'  casts  a  shadow 
on  the  side  of  the  front  brace  at  T',  found  by  drawing  the  ray 
g/rp/  (pro^  y^  According  to  the  construction  of  all  the  other 
points  of  shadow  in  this  problem,  T,  its  other  projection,  is  at 


16  GENERAL   PROBLEMS. 

the  intersection  of  the  projecting  line,  T'T,  with  the  other  pro- 
lection,  AT,  of  the  ray  through  A,  B',  as  it  comes — according  to 
previous  explanation  (Prob.  V.) — when  viewing  the  framing  as 
seen  in  the  left  hand  projection.  AB  being  parallel  to  the  side 
of  the  front  brace,  its  shadow,  tv,  through  T,  is  parallel  to  itself. 
Lastly,  the  brace  hk" — g'y  casts  a  shadow  on  the  side  of  the  post. 
A  little  consideration  of  a  model  will  show,  that,  when  the  brace 
makes  a  less  angle  with  the  horizontal  plane  than  the  light  does, 
its  left  hand  front  edge  and  right  hand  back  edge,  as  seen  on  the 
left  hand  elevation,  will  cast  shadows.  When  the  former  angle 
is  the  greater  of  the  two,  the  right  hand  front  edge  and  left  hand 
back  edge  will  cast  shadows.  When  the  two  above  angles  are 
equal,  the  front  and  back  planes  of  the  brace  will  be  planes  of 
rays,  and  their  traces,  kk'"  and  gh,  produced,  will  bound  the 
shadow  of  the  brace. 

In  the  present  problem,  the  brace  makes  a  less  angle  with  the 
horizontal  plane  than  the  light  does,  hence  the  edges,  k'y — kk" 
and  g'f—hp,  cast  shadows  on  the  post.  The  shadow  of  the  former 
edge  begins  at  k,  and  of  the  latter  at  A,  these  being  the  points 
where  these  edges  pierce  the  post.  Assume  any  point  F,  F7  and 
draw  the  ray  F'H' — FH,  which,  according  to  the  previous  con- 
structions, pierces  the  plane  of  the  side  of  the  post  at  H',H.  H& 
is  therefore  the  shadow  of  kk"  ;  and  sh  parallel  to  it  is  the  shadow 
of  hp. 

PROBLEM  VII. 
To  construct  t/ie  shadow  of  a  circle,  on  tfie  vertical  plane  of  projection. 

Principles. — The  only  essential  difference  between  this  and  the 
previous  problems  is,  that,  in  case  of  a  curve,  a  plane  of  rays  can 
generally  contain  but  one  or  two  points  of  it,  while  it  may  con- 
tain the  whole  of  a  straight  line.  But  we  have  seen  that  the 
shadow  of  a  straight  line  is  practically  determined  by  the  sha- 
dows of  two  of  its  points.  Now  the  shadow  of  a  point  is  found 
in  the  same  way,  whether  that  point  be  on  a  straight  line  or  on 
a  curve  ;  hence  shadows  of  straight  lines,  and  shadows  of  curved 
lines,  are  not  here  distinguished  as  belonging  to  distinct  classes 
of  problems. 

According,  therefore,  to  the  method  of  (30),  again,  we  have 
the  following — 


SHADES  AND  SHADOWS.  17 

Construction. — PI.  III.,  Fig.  10.  Let  the  plane  of  the  circle 
be  perpendicular  to  the  ground  line.  Then  the  equal  lines,  AB 
and  C'D',  perpendicular  to  the  ground  line,  will  be  the  projec- 
tions of  the  circle.  X,et  Aa — AV  be  the  projections  of  a  ray  of 
light.  This  ray  meets  the  vertical  plane  at  a',  the  shadow,  there- 
fore, of  A,  A',  the  foremost  point  of  the  horizontal  diameter  of 
the  circle.  The  ray  B6 — A 'b'  determines  the  shadow  of  the 
point  B,A',  at  b' ;  the  rays  Cc — C'c'  and  Cc — D'df  determine  the 
points  of  shadow,  c'  and  cT,  of  the  extremities  of  the  vertical 
diameter  C — D'C'.  Assuming  either  projection,  as  E,  of  any 
other  two  points  on  the  circle,  their  other  projections  must  be 
constructed  by  some  one  of  the  methods  of  (Prob.  XL.,  Des. 
Geom.),  E  is  the  horizontal  projection,  as  indicated  by  the  con- 
struction, of  the  two  points  E7  and  F'.  Each  of  these,  again,  is 
the  vertical  projection  of  two  points,  of  which  the  two  foremost, 
one  in  each  pair,  are  horizontally  projected  at  G — found  by 
making  CG^CE.  .These  four  points,  G,E';  G,F' ;  E,E' ;  and 
E,F'  cast  shadows  at  e',  g',  nf,  and/',  respectively.  By  joining 
the  points  thus  found,  and  tinting  the  inclosed  figure,  we  shall 
have  the  required  shadow  of  the  circle  on  the  vertical  plane. 

Remarks. — a.  The  lines  a — a! ;  b — b' ;  C' — c',  and  D' — c?',  are 
tangents  to  the  shadow,  which  is  an  ellipse.  Also,  a' — b',  and 
c' — d',  are  conjugate  diameters,  from  which  the  axes  can  be 
found,  if  desired,  as  commonly  shown  among  problems  on  the 
ellipse. 

b.  Observing  that  in  PI.  I.,  Fig.  4,  km  is  the  shadow  of 
a  vertical  edge  at  C  on  the  horizontal  plane  of  projec- 
tion, and  that  in  Plate  III.,  Fig.  10,  a'b'  is  the  shadow  of  the 
diameter  AB — A'  on  the  vertical  plane,  we  conclude  from 
inspection,  "vhat  was  proved  in  (31),  that  whenever  a  line  is 
perpendicular  to  either  plane  of  projection,  its  shadoio  on  that 
plane  is  in  the  direction  of  the  projection  of  a  ray  of  light  on  that 
plane. 

33.  PI.  III.,  Fig.  11,  illustrates  the  forms  of  the  shadows  of  a 
circle,  whose  plane  is  oblique  to  the  direction  of  the  light,  upon 
several  planes  variously  inclined  to  the  plane  of  the  circle. 

AB  represents  a  vertical  projection  of  a  horizontal  circle,  and 
ABGC  the  vertical  projection  of  an  oblique  cylinder  of  rays 
having  the  circle  AB  for  its  base,  and  parallel  to  the  vertical 
plane  of  projection.  The  traces  drawn  through  C,  are  the 
traces  of  a  number  of  planes,  perpendicular  to  the  vertical  plane 

2 


18  GENERAL  PROBLEMS. 

of  projection,  and  whose  intersections  with  the  cylinder  of  rays 
are  the  shadows  of  the  circle  AB. 

DC  is  a  circular  shadow,  equal  to  AB,  its  plane,  DC,  being 
parallel  to  AB.  CF  is  an  elliptical  shadow  less  than  the  circle 
AB,  and  whose  transverse  axis,  projected  at  the  point/  is  equal 
to  the  diameter  of  AB.  CF  being  a  plane  of  right  section,  con- 
tains the  least  shadow  of  AB.  Oil  is  an  elliptical  shadow 
greater  than  AB,  CH,  its  transverse  axis  being  greater  than  AB, 
and  h,  its  conjugate  axis,  being  equal  to  AB.  Between  CF,  the 
least,  and  CH,  the  greater  shadow,  there  must  be  one  equal  to 
the  circle  AB.  This  is  found  by  making  GCF=FCD  when 
CG=CD,  and  as  all  the  diameters  h,  g,f,  etc.,  are  equal,  CG  is 
another  circular  shadow,  called  the  subcontrary  section  of  the 
cylinder,  to  distinguish  it  from  the  parallel  section  CD.  Finally, 
all  shadows  between  the  circular  ones  CD  and  CG,  are  ellipses, 
less  than,  and  all  exterior  to  these  are  ellipses  greater  than  the 
circle  AB. 

34:.  The  foregoing  problems  being  sufficient  to  illustrate  the 
method  of  (30)  we  have  the  following  general  method  for  the 
more  general  case  in  which  the  plane  receiving  the  shadow  is 
not  perpendicular  to  either  plane  of  projection.  Pass  planes  of 
rays,  each  of  which  will  cut  a  point,  p,  from  the  line  L,  casting  the 
shadow,  and  a  line,  k,  from  the  plane,  Q,  receiving  the  shadow. 
The  point  in  which  a  ray  through  the  point,  p,  meets  the  trace,  k, 
will  be  a  point  of  the  shadoio  ofljonQ,. 

Remarks. — a.  When  L  is  a  straight  line,  the  plane  of  rays 
may  always  contain  it,  instead  of  merely  cutting  points  from  it, 
and  then  the  trace  of  the  plane  of  rays  on  the  plane  Q,  will  be 
the  shadow  of  L. 

b.  The  trace  just  mentioned,  must,  however,  be  constructed 
by  the  above  method,  as  seen  in  the  line  cd,  Prob.  III.  Hence, 
practically,  planes  of  rays  are  not  immediately  passed  through 
straight  lines ;  i.  e.  without  construction,  unless  those  lines  are 
perpendicular  to  a  plane  of  projection  ;  in  which  case  the  planes 
of  rays  containing  them  will  be  so  also,  and  their  traces  will  be 
at  once  known.  (29.) 


SHADES  AND  SHADOWS.  19 

PKOBLEM  VIII. 

To  construct  the  shadow  of  a  straight  line  on  a  plane  whose  traces 
are  parallel  to  the  ground  line. 

Construction. — PI.  L,  Fig.  3.  as — a's'  is  the  given  line,  and 
MK  and  M'K'  are  the  traces  of  the  given  plane.  OTO'  is  a 
vertical  auxiliary  plane  which  contains  the  given  line,  as — a's', 
and  intersects  the  given  plane  in  the  line  nT — n'O'.  Hence, 
from  Des.  Geom.  (Prob.  VI.),  the  given  line  intersects  the  given 
plane  at  s',s,  which  (Prob.  III.)  is  one  point  of  the  required 
shadow.  LJI/  is  another  vertical  auxiliary  plane,  containing 
the  ray  ab — a'b',  drawn  through  the  point  a,a'  of  the  given  line 
as — a's'.  This  plane  intersects  the  given  plane  in  the  line  LJ — 
fe'L',  hence  the  ray  through  a,af  intersects  the  given  plane  at 
b',b,  which  by  (34)  is  therefore  another  point  of  the  required 
shadow.  Hence  bs — b's'  is  the  required  shadow. 

Remarks. — a.  The  auxiliary  planes  might  have  been  placed 
perpendicular  to  the  vertical  plane  of  projection.  In  that  case, 
the  horizontal  projections  b  and  s  of  the  points  of  the  shadow 
would  have  been  found  first. 

b.  Let  this  problem  be  solved  thus,  and  also  when  the  given 
plane  has  any  oblique  position. 

PKOBLEM  IX. 

To  find  the  shadow  of  a  chimney,  situated  on  the  end  portion  of  a 
hipped  roof,  upon  the  end  and  side  roofs. 

Principles. — It  only  need  be  noticed  here,  first,  that  as  most 
of  the  edges  of  the  chimney  are  perpendicular  to  one,  or  the 
other,  of  the  planes  of  projection,  the  auxiliary  planes  of  rays 
may  all  contain  edges  of  the  chimney,  and  be  also  perpendicular 
to  one  of  the  planes  of  projection;  and  second,  that  when  a 
plane,  P,  is  perpendicular  to  either  plane  of  projection,  all 
lines  in  that  plane,  P,  will  be  projected  on  the  plane  of  projec- 
tion, to  which  it  is  perpendicular,  in  its  trace  on  that  plane 
(29). 

Construction.— PI.  I.,  Fig.  5.  URR"T— K'T'  is  the  horizon- 
tal plane  of  the  eave-trough,  intersected  by  the  roof  in  the  lines 


20  GENERAL   PROBLEMS. 

TE,  EG,  and  GU.  GIU— GTT'  is  the  front  roof,  EIG— GT 
the  end  roof,  perpendicular  to  the  vertical  plane  of  projection, 
and  EIT — GTT'  is  the  back  roof;  xw  is  the  chimney-flue, 
dve  a  horizontal  section  of  its  outside  surface,  and  ABD — A'C'D' 
is  its  abacus. 

The  shadow  of  the  abacus  on  the  front  of  the  chimney,  ia 
found  as  in  Prob.  IY. 

To  find  the  shadows  on  the  roof:  First,  the  shadow  on  the 
front  roof.  C'Y'  is  the  vertical  trace  of  a  plane  of  rays  through 
EC — C',  and  perpendicular  to  the  vertical  plane  of  projection. 
According  to  principle  second,  O'Y'  is  the  vertical  projection  of 
its  trace  on  the  front  roof.  Y' — Y"Y  is  its  trace  on  the 
horizontal  plane  of  the  eave-trough  ;  hence,  projecting  0'  at  O 
we  have  OY  and  OY",  for  the  horizontal  projections  of  its 
traces  on  the  front  and  back  roofs.  Drawing  the  ray  BA — 
whose  vertical  projection  is  C'Y' — and  projecting  h  at  h',  the 
point  A,  h'  is  the  shadow  of  B,C',  and  hO — /I'O',  the  shadow  of 
BC — C'  on  the  front  roof.  Drawing  a  similar  plane  of  rays, 
BT/,  its  traces  on  the  roof  are  F'L'— FL",  parallel  to  O'Y'- 
OY",  a  line  from  L  parallel  to  YO,  and  the  short  line,  paral- 
lel to  GE,  across  the  end  roof  at  F.  The  ray  B^ — B'L'  meets 
the  trace  FL"  at  g,g',  giving  hg — li'g'  for  the  shadow  of  B — B'C'. 
From  g,gf  the  shadow  of  AB — A'B',  on  the  parallel  front  roof, 
is  the  parallel  line,  gf—g'f.  Drawing  the  ray  /K— /'K',  we 
find  the  portion  BK — B'K'  whose  shadow  is  gf—g'f. 

Second. — The  shadow  on  the  end  roof,  ec  is  the  trace,  on  this 
roof,  of  a  vertical  plane  of  rays  through  the  vertical  edge,  e — 
e"e',  of  the  chimney.  This  plane  cuts  the  edge,  AB — A'B',  of 
the  abacus  in  the  point  a,a',  through  which,  drawing  the  ray 
a'c' — oc,  and  projecting  c'  at  c,  we  find  ec,  the  shadow  of  the 
portion  e — e"e'  of  the  edge  of  the  chimney.  Above  e'  this  edge 
is  in  the  shadow  of  the  abacus  on  the  chimney.  Joining  c,c' 
and  f,/';  cf—c'f  is  the  shadow  of  aK — a'K'.  Likewise,  du  is  the 
shadow  of  the  vertical  edge  at  d.  By  projecting  u  at  u',  and  by 
drawing  the  vertical  projection  of  a  ray  through  u',  we  could 
find  the  portion  of  the  edge  from  d,d'j  upwards,  which  casts  the 
shadow  du — d'u'. 

Third. — To  find  the  shadow  on  the  back  roof  and  eaves  plane. 
A'o'  is  the  vertical  trace  of  a  plane  of  rays  parallel  to  the  plane 
C'Y'Y".  Its  trace  on  the  back  roof  is  NV — oN.  It  intersects 
the  edge,  d — d'v',  of  the  chimney  at  d,v',  through  which,  draw- 


SHADES  AND  SHADOWS.  2l 

ing  a  ray,  v'o' — dp,  we  find'  pjp'  the  limit  of  the  shadow  of  d — 
d'v'  on  the  back  roof.  This  point  p,p'  is  also  by  construction 
the  shadow  of  r,A'  where  the  ray  p'v' — pd  meets  the  edge 
AD — A'  of  the  abacus.  Drawing  the  ray  or",  rr"  is  determined 
as  the  portion  of  this  edge,  which  casts  the  shadow  po — p'o'. 
The  remaining  portion,  Dr",  casts  the  parallel  shadow  on — o',  a 
part  of  the  trace  of  the  plane  of  rays  AV  on  the  plane  of  the 
eave-trough,  and  limited  by  the  ray  Dn — AV.  Drawing  the 
ray  Y6,  we  find  the  limit  of  that  portion  of  the  edge  BC — C' 
which  casts  a  shadow  on  the  back  roof.  &C  casts  the  equal  and 
parallel  shadow,  Yy,  on  the  parallel  plane  of  the  eave-trough. 
Likewise  ym  is  equal  and  parallel  to  CD — C'D',  and  is  limited 
by  the  rays  Cy  and  Dm.  Then  mn  is  the  shadow  of  D — A'D', 
which  completes  all  the  required  shadows. 

Remarks. — a.  The  boundary  of  the  shadow  on  the  back  roof, 
as  seen  in  vertical  projection,  is  fringed,  to  give  it  greater  dis- 
tinctness. 

b.  To  locate  either  projection  of  a  point  on  the  side  roofs,  one 
projection  of  such  a  point  being  given.     Let  H  be  the  horizontal 
projection  of  a  vertical  rod  on  the  front  roof.     Draw  Hg  parallel 
to  the  trace,  GU,  of  the  front  roof;  q  is  vertically  projected  at 
q1,  and  j'H',  parallel  to  GT',  is  the  vertical  projection  of  gH ; 
hence  H'H"  is  the  vertical  projection  of  the  rod  at  H.     This 
construction  would  be  used   in  locating  the  intersection,  with 
either  front  or  back  roof,  of  a  chimney  situated  on  either  of  those 
roofs.     (See  Des.  Geom.  Prob.  X.) 

c.  To  find  the  shadow  of  the  rod,  H — H'H",  pass  a  vertical 
plane  of  rays  through  it ;  HJ,  parallel  to  ~Bg,  is  the  horizontal 
projection  of  the  trace  of  this  plane  on  the  front  roof.     Project 
J  at  J'  and  H'J'  will   be  the  vertical   projection  of  the  same 
trace.     The  ray  H"A'" — Kh"  meets  this  trace  at  h"',h",  giving 
HA"-— TL'h'"  for  the  projections  of  the  shadow  of  the  rod.     This 
construction  could  have  been  employed  in  finding  all  points  of 
the  shadow  of  the  chimney,  instead  of  making  the  planes  of 
rays  perpendicular  to  the  vertical  plane.     Observe  that  in  the 
construction  given  for  the  chimney,  we  find  the  horizontal  pro- 
jections of  points,  as  h,  first ;  while  in  the  construction  of  h"h'" 
we  find  h"f  first. 

35.  In  dismissing  the  subject  of  shadows  on  planes,  it  may 
be  noticed,  that  it  is  sometimes  advantageous  to  consider  the 
ray  of  light,  itself,  as  making  an  angle  of  45°  with  one  of  the. 


22  GENERAL  PEOBLEMS. 

planes  of  projection.  For,  suppose  such  a  ray  to  be  passed 
through  the  uppermost  point  of  a  vertical  line,  and  to  make  an 
angle  of  45°  with  the  horizontal  plane.  Evidently,  the  distance 
from  the  foot  of  the  line  to  the  intersection  of  the  ray  with  the 
horizontal  plane — which  would  be  the  shadow  of  the  line — 
would  then  be  equal  to  the  height  of  the  line.  Hence,  without 
having  any  vertical  projection  of  the  line,  its  height  may  be 
known  from  its  shadow.  Or,  which  amounts  to  the  same  thing, 
the  height  of  a  point  above  the  horizontal  plane  will  be  known 
by  the  distance  of  its  shadow  from  its  horizontal  projection. 

The  same  essential  principle  is  true  in  more  general  terms. 
For,  by  knowing  the  direction  of  the  light,  we  can  find  the  rela- 
tive lengths  of  vertical  lines  and  their  shadows,  and  this  relation 
being  known,  we  can  find  the  height  of  all  points  of  an  object, 
having  given  only  its  plan  and  shadow.  The  same  principles 
are  applicable  to  constructions  made  on  the  vertical  plane  of 
projection.  To  illustrate,  let  us  take  the  following  example  : 


PEOBLEM  X. 
To  find  the  shadow  of  a  shelf  and  brackets  upon  a  vertical  plane. 

Construction. — PL  II.,  Fig.  9.  Let  AB,  taken  as  the  ground 
line,  be  the  horizontal  trace  of  the  given  vertical  surface ;  let 
CD — C'D'  be  the  lower  front  edge  of  the  shelf,  and  let  cl — c'V 
be  one  of  the  brackets. 

Let  E'F  be  the  vertical  projection  of  a  ray,  drawn  in  any 
direction,  relative  to  the  ground  line.  It  is  sufficient  to  under- 
stand that  this  ray  makes  an  angle  of  45°  with  the  vertical 
plane.  Then  the  shadow,  c'd',  of  the  edge  en — c'  will  be  parallel 
to  E'F  and  equal  to  en.  Next,  d'e'  can  be  drawn,  parallel  to 
ca — cV,  and  limited  by  the  ray  a'e',  and  the  shadow  of  the 
unshaded  part  of  a — a'V  will  be  the  parallel  line  e'k'. 

Likewise,  taking  any  point  //,  in  the  edge  CD — C'D'  of  the 
shelf,  make  the  ray  //*'  equal  to  fg:  then  h'k'}  parallel  to  C'D'f 
will  be  the  shadow  of  CD— C'D'.  " 

Remark. — Though  not  necessary,  the  horizontal  projection  of 
the  ray  may  be  found  as  follows:  Make  GA=E'F,  then  will 
GF — E'F  be  the  two  projections  of  a  ray  which  makes  an  angle 
of  45°  with  the  vertical  plane  of  projection. 


SHADES  AND   SHADOWS.  23 


CLASS  II. 

SHADES  AND   SHADOWS  ON  SINGLE   CURVED   SURFACES,   IN 
GENERAL. 

SECTION  I. 
On  Developable  Single  Curved  Surfaces. 

§  1. — Shades. 

36.  The  line   of  shade  on  a   developable  surface,  consists  of 
those  reclilineUr  elements  along  which  planes  of  rays  are  tangent 
to   the  surface,  since  such  a  line  would  evidently  separate  the 
illuminated  from  the  dark  portion  of  the   surface.     Since  the 
tangent  planes   are  planes  of  rays,  they  may  be   regarded  as 
parallel  to  a  given  ray,  in  space.     Hence  the  general  method  for 
the  construction  of  the  elements  of  shade  on  a  developable  sur- 
face, resolves  itself  into  the  operations  of  (Des.  Geom.  Probs. 
53,  &c.)     "  To  construct  a  plane,  parallel  to  a  given  line,  and  tan- 
gent to  a  developable  single  curved  surface"    Since  two  such  planes 
can  generally  be  found  in  any  case,  there  will  usually  be  two 
opposite  elements  of  shade. 

37.  When  the  indefinite  surface  is  limited  by  one  or  more 
plane  intersections,  taken  as  bases,  the  complete  line  of  shade 
will  consist  of  the  elements  of  shade  on  the  convex  surface,  toge- 
ther with  those  portions  of  the  circumferences  of  the  bases  which 
divide  light  from  dark  portions  of  the  entire  body. 

38.  If  a  plane  of  rays  be  tangent  to  a  developable  surface 
along  one  of  its  elements,  then  any  secant  plane  will  cut  from  the 
surface  a  curve,  and  from  the  plane  of  rays,  a  line,  tangent  to  the 
curve  at  one  point  of  the  element  of  shade.  Moreover,  if  the  secant 
plane  be  perpendicular  to  the  tangent  plane  of  rays,  the  line  cut 
by  it  from  the  latter  will  evidently  be  the  projection  of  the  light 
upon  the  secant  plane,  since  the  tangent  plane,  when  thus  per- 
pendicular to  the  secant  plane,  becomes  a  projecting  plane  of 
rays  upon  the  latter  plane. 


24  GENERAL   PROBLEMS. 

PROBLEM  XL 
To  consiruct  ike  elements  of  shade  on  any  cylinder  or  cone. 

Constructions. — First. — -To  construct  the  element  of  shade  on  a 
horizontal  right  semi- cylinder.  PI.  IV.,  Fig.  16.  Let  Yb  be  the 
trace  of  one  of  two  vertical  planes  of  projection — upon  the  plane 
of  the  paper — taken  as  the  second  vertical  plane,  and  at  right 
angles  to  the  former  one.  Let  the  former  plane  be  perpendicular 
to  the  common  axis  of  the  cylinder,  and  cylindrical  abacus,  shown 
in  the  figure,  and  let  the  latter  plane  contain  the  axis  of  the 
cylinder.  The  projections  will  then  be  as  in  the  figure.  Let 
De — DY  be  the  projections  of  a  luminous  ray.  De  is  also  the 
trace,  on  the  right  hand  plane,  of  a  plane  of  rays  perpendicular 
to  that  plane ;  e  is  therefore  the  right  hand  elevation  of  its 
element  of  contact  with  the  cylinder,  that  is,  of  the  element  of 
shade  of  the  cylinder.  E'e"  is  therefore  the  left,  or  linear  elevation 
of  the  same  element. 

Likewise,  the  parallel  tangent  plane  of  rays  at  G,  determines 
the  element  of  shade,  G — G^G',  on  the  abacus. 

/Second. — To  construct  the  elements  of  shade  on  a  cylinder  whose 
axis  is  oblique  to  both  planes  of  projection.  PI.  III.,  Fig.  13.  ABF 
— A'G'  is  the  lower  base  of  the  cylinder.  Ibe — FJ'  is  its  upper 
base,  Cb  and  M</  are  its  extreme  elements,  as  seen  in  horizontal 
projection,  and  A'F  and  G'J',  the  extreme  elements  as  seen  in 
vertical  projection.  The  axis  of  the  cylinder  pierces  the  hori- 
zontal plane  at  K,  the  centre  of  the  lower  base.  The  ray  «L — 
a'L',  through  a  point  a,a'  of  the  axis,  pierces  the  same  plane  at 
L,  hence  KL  is  the  horizontal  trace  of  a  plane  containing  the  axis 
and  a  ray.  From  (5)  such  a  plane  is  a  plane  of  rays  through 
the  axis.  The  tangent  planes  of  rays  will  be  parallel  to  the  one 
just  found,  hence  their  horizontal  traces  will  be  parallel  to  KL. 
BB"  and  FF"  are  these  traces,  and  their  points  of  tangency,  B 
and  F,  with  the  lower  base,  are  where  their  elements  of  contact 
— the  elements  of  shade — pierce  the  horizontal  plane.  Hence  Bw 
— BV,  and  F/*—  F/',  are  the  required  elements  of  shade;  one 
projection  of  each  being  visible. 

Tliird. — To  construct  t/ie  elements  of  shade  on  a  cylinder  whose 
axis  is  horizontal,  but  oblique  to  the  vertical  plane  of  projection.  PI. 
IV.,  Fig.  17.  In  order  to  construct  the  vertical  projection  of  a 
cylinder,  thus  seen  obliquely,  it  is  necessary  to  have,  first,  its 


SHADES  AND  SHADOWS.  25 

projection  on  a  plane  perpendicular  to  its  axis.  Then  let  NZH 
be  the  horizontal  projection  of  the  cylinder,  and  take  O"C", 
perpendicular  to  its  axis,  for  the  ground  line  of  a  new  vertical 
plane,  parallel  to  the  base  NZ.  This  base  will  then  have  the 
circle  H"O"6"  for  its  auxiliary  vertical  projection,  and  that  of  theVj 
cylinder.  Now,  any  point  as  Z',  in  the  principal  elevation,  is  inj 
a  projecting  line,  ZZ',  and  at  a  height,  Z'O',  from  the  ground 
line,  equal  to  the  height  of  the  same  point,  as  seen  at  Z",  above 
the  ground  line  O"C". 

Having  thus  found  the  projections  of  the  cylinder,  we  may- 
assume  any  two  projections  of  a  luminous  ray,  and  find  the  third. 
Let  FL — F'L'  be  the  assumed  projections  of  a  ray.  L,L'  is  pro- 
jected on  the  auxiliary  plane  at  L"; — I/,  and  L"  being  at  equal 
heights  above  their  respective  ground  lines.  Likewise  G"  and 
F',  both  projected  from  F,  are  at  equal  heights  above  their 
respective  ground  lines.  GK'L"  is  then  the  auxiliary  projection 
of  the  ray.  "  Hence  also  T"H"  and  M"J",  parallel  to  G"L",  are 
the  traces,  on  the  auxiliary  vertical  plane,  of  tangent  planes  of 
rays  perpendicular  to  that  plane.  Hence  H" — HI  and  J" — KJ 
are  two  projections  of  the  two  elements  of  shade  of  the  cylinder. 
KJ  is  again  vertically  projected  at  K'J'  at  a  height,  equal  to 
J'V,  above  the  ground  line  C'O',  or  by  projecting  J  into  the 
principal  vertical  projection  of  the  base  at  J',  and  drawing  J'K' 
parallel  to  the  ground  line.  The  former  construction  is  obviously 
more  accurate  in  practice. 

Otherwise :  omitting  all  the  operations  of  the  last  paragraph, 
the  ray,  NO — N'O',  may  be  projected  into  the  plane,  NZ,  of  the 
base  of  the  cylinder,  by  projecting  0,0'  at  o",o'".  Then,  con- 
sidering that  N,N'  is  its  own  projection  on  the  plane  NZ,  we 
have  No" — Wo'"  to  represent  the  projections  of  the  ray  on  this 
plane.  Hence  draw  a  line  tangent  to  the  base,  and  parallel  to 
NV",  and  by  the  principle  of  (38)  J'  will  be  its  point  of  tan- 
gency  with  the  base  N'J'Z',  and  therefore  a  point  of  the  element 
of  shade  J'K'. 

Fourth. — To  construct  the  elements  of  shade  on  a  cone.  PI.  III., 
Fig.  12,  and  PI.  V.,  Fig.  18. 

Principles. — A  plane  is  tangent  to  a  cone  along  an  element, 
and  all  the  elements  contain  the  vertex  of  the  cone,  hence  a  plane 
of  rays  tangent  to  a  cone  will  contain  the  ray  through  its  ver- 
tex. Therefore  the  trace  of  this  plane  on  any  given  plane,  will 
contain  the  intersection  of  that  ray  with  this  given  plane,  and, 


26  GENERAL  PROBLEMS. 

by  (38)  the  point  of  contact  of  this  trace  with  the  section  of  the 
cone,  made  by  the  given  plane,  will  be  a  point  of  the  element  of 
shade. 

Constructions.— In  PI.  III.,  Fig.  12,  draw  a  ray,  VE— Y'E', 
from  the  vertex,  V,  V,  of  the  cone.  It  pierces  the  horizontal 
plane,  which  is  here  the  plane  of  the  cone's  base,  at  E.  Then 
ET  is  the  horizontal  trace  of  one  of  the  two  tangent  planes  which 
can  be  drawn,  and  TV — T'V  is  the  element  of  shade  determined 
by  it.  In  PI.  V.,  Fig.  18,  as  the  ray,  YN— Y'E'  pierces  the 
horizontal  plane  within  the  cone's  base,  on  that  plane,  no  lines 
can  be  drawn  from  N,  tangent  to  the  base.  This  indicates  that 
no  planes  of  rays  can  be  drawn  tangent  to  the  cone.  The  lower 
nappe,  therefore,  would  be  wholly  illuminated,  but. for  the  pre- 
sence of  the  upper  nappe,  which  casts  a  shadow  upon  it. 

Fifth. — To  construct  the  elements  of  shade  on  two  intersecting 
cylinders,  wJiose  aoces  are  at  right  angles  to  each  other,  in  a  plane 
parallel  to  the  vertical  plane  of  projection,  one  of  these  axes  being  ver- 
tical. PI.  V.,  Fig.  19.  The  elements  of  shade  on  the  vertical 
cylinder  are  found  at  once,  by  drawing  twTo  vertical  planes  of 
rays,  tangent  to  that  cylinder.  A/  is  the  horizontal  trace  of 
such  a  plane,  coinciding  with  the  horizontal  projection  of  a  ray, 
and  giving  the  element  of  shade  A — A'A". 

The  elements  of  shade  on  the  horizontal  cylinder,  are  found 
by  revolving  the  plane  of  either  of  its  bases  either  into,  or  paral- 
lel to,  one  of  the  planes  of  projection.  Let  the  plane  uDv '  be 
revolved  about  its  vertical  trace  Di/,  into  the  vertical  plane  of 
projection.  The  centre  n,ri  of  the  base  contained  in  this  plane, 
will  then  revolve  in  the  arc  nn'" — n'n",  to  n".  The  circle,  with 
n"  as  a  centre,  and  radius  n"T"=n'E',  will  then  be  the  revolved 
position  of  the  base  uy — eV,  and  the  projection  of  the  entire 
horizontal  cylinder  upon  the  plane  uT)v'.  Now,  T"7c"  is  the 
trace  on  the  plane  of  the  revolved  base,  of  a  plane  of  rays  per- 
pendicular to  that  plane  and  tangent  to  the  horizontal  cylinder, 
T"&"  being  drawn  in  any  assumed  direction,  if  the  principal  pro- 
jections of  a  ray  are  not  given.  T",  and  the  point  V"  diametri- 
cally opposite,  are  then  the  auxiliary  projections  of  the  elements 
of  shade  of  the  horizontal  cylinder.  By  a  counter  revolution,  in 
which  these  points,  considered  as  the  extremities  of  these  ele- 
ments of  shade,  revolve  in  the  arcs  T'Y — ,  t"e  and  Y'V — v"p, 
we  find  ell — e'W  for  the  lower  front  element  of  shade,  and 
pV — v'V  for  the  upper  back  element  of  shade. 


SHADES  AND  SHADOWS.  27 

Remark. — If  the  principal  projections  of  the  light  were  given, 
T"&",  instead  of  being  assumed,  would  be  constructed  as  the 
auxiliary  projection  of  the  light  upon  the  plane  uDv',  by  an 
obvious  construction. 


§  n. — Shadows. 

39.  As  in  the  case  of  planes,  so  in  that  of  single  curved  sur- 
faces, when  all  their  elements  are  parallel,  the  projection  of  the 
surface  will  be  a  line,  when  it  is  perpendicular  to  a  plane  of  pro- 
jection.    The  projection  of  the  perpendicular  plane  will,  how- 
ever, be  a  straight  line,  while  that  of  the  single  curved  surface 
will  be  a  curve  ;  viz.  its  curve  of  right  section.     As  cylinders 
alone,  among  single  curved  surfaces,  have  parallel  elements,  they 
only  can  have  all  their  elements  perpendicular  to  the  same  plane, 
and  hence  they  only  can  be  projected  in  a  line. 

40.  For  all  cases  of  shadows  upon  cylinders  whose  axes  are 
perpendicular  to  a  plane  of  projection,  we  have,  from  the  pre- 
ceding article,  the  following  special  method.     Determine,  by  inspec- 
tion, the  intersection,  M,  of  a  ray  through  any  point,  P,  of  the  line 
casting  the  shadow,  with  the  linear  projection  of  the  cylinder.     This 
will  be  one  projection  of  the  shadow  of  P.     Its  other  projection  will 
be  Hie  intersection  of  the  line  perpendicular  to  the  ground  line,  Hirough 
M,  with  the  otiier  projection  of  the  ray  through  f. 

The  two  following  problems  illustrate  this  method. 


PROBLEM  XII. 


To  construct  the  shadow  of  the  projecting  head  of  a  cylinder  upon 

the  cylinder. 

PI.  IY.,  Fig.  16.  The  projections  of  this  body  have  been  fully 
described  in  the  first  case  of  the  preceding  problem.  The  right- 
hand  edge,  G"A',  of  the  head,  above  the  element  of  shade,  G — 
G'G",  casts  the  shadow  required.  Drawing  the  projection,  Cc?, 
of  a  ray,  on  the  plane  containing  the  linear  projection,  bdE,  of 
the  cylinder,  d  is  at  once  found,  as  one  projection  of  the  shadow 
of  C,C'  upon  the  cylinder.  Its  other  projection,  d',  is  the  inter- 
section of  the  perpendicular,  d — df,  to  the  trace  (ground  line)  F6, 
•with  the  other  projection,  C'ef,  of  the  ray  through  0,0'.  The 


28  GENERAL   PROBLEMS. 

other  points  of  shadow  are  found  in  exactly  the  same  way,  as  is 
evident  by  inspection.  The  tangent  ray,  De — D'e',  determines 
that  point  of  shadow,  e,e',  in  which  the  shadow  is  lost  in  the 
element  of  shade,  e — e'e". 


PROBLEM  XIII. 

To  construct  the  shadows  of  the  edges  of  shade  of  a  cross  upon  a 
cylinder,  both  bodies  being  seen  obliquely. 

Principles. — PI.  IY.,  Fig.  17.  The  three  projections  of  the 
cross  and  cylinder,  of  the  ray  of  light,  and  of  the  elements  of 
shade  of  the  cylinder,  having  been  already  constructed,  as 
explained  in  Prob.  XI.  (Third),  we  may  proceed  at  once  with 
the  construction  of  the  shadow,  after  having,  first,  determined, 
by  inspection,  the  edges  of  shade  of  the  cross,  and,  second,  the 
limits  of  that  part  of  the  body  of  the  cross,  whose  edges  of  shade 
cast  shadows  on  the  cylinder. 

The  edges  of  shade  must,  evidently,  here  be  determined  with 
reference  to  a  single  direction  of  the  luminous  ray  in  space;  that 
is,  we  must  find  the  projections  of  those  edges,  on  the  three 
planes  of  projection,  which  in  space  really  cast  shadows,  when, 
the  light  comes  in  the  single  direction  projected  in  FL — FT/ — 
Gr"L".  These  edge*  are,  in  the  present  case,  B"D"— BD— B'D' ; 
the  one  diagonally  opposite,  through  YY";  then  c" — c"'P — c'P', 
and  the  one  diagonally  opposite,  G"— FG— F'G' ;  also  X"G"— 
XG — X'G' ;  its  diagonally  opposite  edge,  c"w" — c'"w ;  next 
XV— XP— XT' ;  and  its  diagonally  opposite  edge,  Fw— G"*, 
and  the  horizontal  edge  at  D",D,D'. 

The  limits  of  that  portion  of  the  cross,  which  casts  a  shadow 
on  the  cylinder,  are  found  by  drawing  the  two  tangent  planes  of 
rays,  whose  traces  are  IVT'J"  and  T"H",  and  which  determine 
the  elements  of  shade  on  the  cylinder  (Prob.  XI.  Third).  These 
planes  contain  those  points  of  the  cross,  which  cast  shadows  on 
the  elements  of  shade,  where  the  shadow  disappears  from  the 
cylinder. 

Construction. — The  point  M",M,M',  cut  from  the  cross  by  the 
plane  M"J",  casts  a  shadow  on  the  element  of  shade,  at  a  point 
whose  auxiliary  projection  is  J",  whose  horizontal  projection 
may  be  found  by  projecting  3"  into  the  horizontal  projection, 
Ma,  of  the  ray  through  M",M,  at  a ;  and  whose  vertical  pro- 


SHADES  AND  SHADOWS.  29 

jection  may  then  be  found  by  projecting  a  into  the  vertical  pro- 
jection, MV,  of  the  same  ray,  at  a'.  Otherwise :  we  may  con- 
struct the  vertical  projection,  J'K',  of  the  element  of  shade,  on 
which  the  shadow  of  M",M'  is  known  to  fall,  as  explained  in 
(Prob.  XI.  Third}.  Then  the  intersection,  a',  of  this  element 
with  the  ray,  vertically  projected  in  MV,  is  the  vertical  projec- 
tion of  the  required  point  of  shadow,  which  may  be  then  pro- 
jected into  the  horizontal  projection,  Ma,  of  the  same  ray, 
at  a. 

Thus  every  point  of  shadow  may  be  found,  first  in  horizontal 
projection,  and  then  projected  up ;  or  first  in  vertical  projection, 
and  then  projected  down.  Also,  a  may  be  projected  up  into 
J'K',  instead  of  into  M'a' ;  or,  a'  being  first  found,  as  above,  it 
may  be  projected  down  into  JK,  to  find  a.  Indeed  this  is.  prac- 
tically, the  most  exact  construction. 

As  all  the  other  points  of  the  shadow  are  found  precisely  as 
just  described,  we  shall  only  point  out  those,  which,  in  the 
present  figure,  it  is  desirable  to  find.  This  will  assist  in  solving 
the  problem,  when  the  relative  position  of  the  bodies,  the  light, 
and  the  planes  of  projection,  are  slightly  changed. 

The  shadow  on  the  foremost  element,  b" — vN — v'X',  is  cast 
by  the  point,  R",  R,R',  in  the  ray  &"R",  and  falls  at  &",&,&'. 

In  the  figure,  the  cross  is  made  tangent  to  the  cylinder  along 
the  edge  c" — c'"P,  which  is  therefore  its  own  shadow.  The 
highest  point,  S",  of  the  edge  B"X",  which  casts  a  shadow,  is  found 
by  drawing  the  ray  c"S".  Its  shadow  is  c",c,c'.  The  frag- 
ment, S"X",  casts  the  shadow,  cT,  on  the  front  of  the  short  arm 
of  the  cross. 

The  shadow  on  the  highest  element,  is  found  by  drawing  the 
ray  p"Q,'f,  which  determines  the  point  Q",Q,Q',  whose  shadow 
is  p"&e'.  The  edge  XV— XP  casts  the  shadow  Pd— P'd'  • 
the  edge  XGr,  the  shadow  from  d  to  the  point  just  beyond  e ;  the 
edge  GF,  the  shadow  from  the  latter  point  to  o ;  the  edge  through 
F  and  w,  the  shadow  op,  and  the  edge  through  c"  and  w",  the 
shadow  pc'",  partly  invisible.  The  portion  G"T"— UT,  of  UD, 
casts  the  shadow  gf,  and  the  portion,  V"^  of  the  lower  left  hand 
edge,  the  shadow  Teh. 

JfemarJcs. — a.  The  portion  of  shadow  cast  on  the  horizontal 
plane  in  front  of  the  cylinder  is  shown.  It  is  found  as  in  Prob. 
III.  The  shadow  of  the  cylinder,  and  of  the  part  of  the  cross 
which  is  beyond  the  plane  H"T",  can  be  found  on  the  horizontal 


30  GENERAL    PROBLEMS. 

plane  in  the  same  manner,  and  will  add  to  the  beauty  of  the 
figure,  especially  when  shaded  in  graduated  tints. 

b.  If  the  light  were  to  make  a  greater  angle  with  the  horizon- 
tal plane  than  the  edges  parallel  to  X"c",  as  seen  in  the  auxiliary 
projection,  the  edges,  as  X" — XT,  would  be  edges  of  shade. 

c.  Those  edges  are  inked  as  lines  of  shade,  in  the  auxiliary 
elevation,  which  are  so,  in  reference  to  the  single  direction  of  the 
luminous  ray,  shown  in  this  figure. 

41.  The  general  method  for  rinding  shadows  on  oblique  cylin- 
ders, and  other  single  curved  surfaces,  is  this :  Pass  planes  of 
rays  so  that  each  shall  cut  the  line  casting  the  shadoiv  in  a  point  P, 
and  the  single  curved  surface  in  a  straight  element,  E — or  in  some 
other  simple  section.  The  point,  S,  where  tJie  ray  through  P  meets 
the  element,  E,  is  tJie  shadow  of  P  upon  E,  and  is  therefore  one  point 
of  Oie  shadow  of  the  given  line  upon  the  given  surface. 

Remarks. — a.  For  the  cylinder;  a  plane  of  rays  must  be 
parallel  to  the  axis,  in  order  to  intersect  the  cylinder  in  straight 
elements. 

b.  For  the  cone ;  a  plane  must  contain  the  ray  through  the 
vertex,  in  order  to  intersect  it  in  straight  elements. 

c.  The  two  following  problems  illustrate  this  general  method. 


PROBLEM  XIY. 

To  construct  the  shadow  cast  by  the  upper  base  of  a  hollow  oblique 
cylinder  upon  its  interior. 

Principles. — The  method  of  (41)  applied  to  this  problem  gives 
the  following  principles  of  solution.  Any  secant  plane,  parallel 
to  the  plane  containing  the  axis  and  a  ray,  will  cut  two  recti- 
linear elements  from  the  cylinder,  one  of  which  will  be  towards 
the  source  of  light.  The  upper  extremity  of  this  element  casts  a 
point  of  shadow  on  the  opposite  element  in  the  same  plane.  The 
elements  contained  in  the  same  tangent  planes  of  rays,  coincide 
in  one,  which  is  the  element  of  shade.  Hence  the  upper  extre- 
mity of  this  element  is  also  a  point  of  shadow  on  the  interior  of 
the  cylinder. 

Constructions. — PI.  HI.,  Fig.  13.  According  to  the  foregoing 
principles,  f,f,  and  u,u' ;  the  upper  extremities  of  the  elements 


SHADES  AND  SHADOWS.  31 

of  shade  (Prob.  XL  Second)  are  the  points  where  the  required 
shadow  on  the  interior  begins. 

To  find  any  other  point  of  this  shadow.  The  plane  of  rays 
whose  horizontal  trace,  MET,  is  parallel  to  the  horizontal  trace, 
NEL,  of  the  plane  of  rays  through  the  axis,  contains  the  elements 
M<7  and  He.  The  point,  g,g',  of  the  former  element,  casts  its 
shadow  on  the  latter  element,  at  h,h.  This  point  may  be  found 
by  drawing  the  horizontal  projection,  gh,  of  a  ray,  noting  the 
point  h,  and  then  projecting  it,  either  into  the  vertical  projection, 
g'h',  of  the  ray,  or  into  the  vertical  projection,  D  V,  of  the  ele- 
ment containing  it.  Otherwise  :  the  same  point  may  be  found 
by  first  drawing  the  vertical  projection,  g'h',  of  the  ray  through 
g,g\  noting  its  intersection,  /?',  with  the  vertical  projection,  DV, 
of  the  element  containing  it,  and  then  projecting  it  either  into 
gh,  the  horizontal  projection  of  the  same  ray,  or  into  He,  the  hori- 
zontal projection  of  the  element  containing  h,hr. 

To  find  the  shadow  cast  by  any  particular  point,  assumed  in 
advance,  on  the  upper  base,  as  n,n'.  Draw  the  element,  n — N, 
containing  this  point,  and  through  N,  its  intersection  with  the 
horizontal  plane,  draw  the  trace,  NE,  of  a  plane  of  rays,  and 
then  proceed  as  before,  to  find  0,0',  the  shadow  of  n,nf. 

Conversely,  to  find  the  shadow  cast  upon  any  element,  assumed 
in  advance,  as  Cb — C'J'.  Through  the  intersection,  C,  of  this 
element,  with  the  horizontal  plane,  draw  the  trace  CE,  parallel 
to  NE,  of  the  plane  of  rays  which  cuts  from  the  cylinder  the  ele- 
ment Er,  towards  the  source  of  light ;  whose  upper  extremity,  r, 
casts  the  shadow,  5,5',  on  the  assumed  element  bG — J'C'. 

The  lowest  point,  0,0',  of  the  shade,  is  evidently  on  the  ele- 
ment, EC? — E'cT,  contained  in  the  plane  of  rays  through  the  axis, 
because  the  two  elements  contained  in  this  plane  are  the  farthest 
apart,  being  diametrically  opposite. 

Remarks. — a.  It  does  not  strictly  belong  to  this  problem  to 
construct  the  shadow  of  the  cylinder  on  the  horizontal  plane.  It 
is  however  shown  in  the  figure,  and  is  bounded  by  BB",  the 
shadow  of  the  element  of  shade,  Eu — B  V  ;  the  semi-circle,  B"F", 
whose  centre  is  L,  and  which  is  the  shadow  of  udf,  and  FF",  the 
shadow  of  the  element  of  shade,  Ff—Ff. 

b.  The  vertical  projection  of  the  shadow  on  the  interior  of  the 
cylinder  is  invisible. 

c.  It  happens  in  the  present  figure,  that  the  horizontal  pro- 
jection,/ma,  of  the  same  shadow  is  a  straight  line.     This  shows 


32  GENERAL   PROBLEMS. 

that  tliis  shadow  is  a  plane  curve,  whose  plane  happens,  in  this 
instance,  to  be  vertical.  The  shadow  being  a  plane  curve,  and 
also  the  intersection  of  two  cylinders  having  a  common  base, 
/JI — JT, — viz.  the  given  cylinder,  and  the  cylinder  of  rays 
through  its  upper  base — we  infer,  what  is  actually  true,  that  the 
intersections  of  such  cylinders,  having,  as  they  do,  two  common 
tangent  planes,  are  plane  curves. 

d.  The  last  remark  serves  as  a  simple  illustration  of  the  use  of 
the  exact  constructions  of  Descriptive  Geometry  as  a  means  of 
research  in  discovering  theorems,  particularly  of  form  and  posi- 
tion, relating  to  geometrical  magnitudes. 

42.  In  the  following  problem,  besides  an  illustration  of  the 
general  method  of  (41)  there  is  an  illustration  of  the  special  method 
called  the  method  of  one  auxiliary  shadow,  which  is  as  follows: 

Where  the  shadow  of  a  line  on  any  surface  meets  the  intersec- 
tion of  that  surface  with  a  second  surface,  is  a  point  of  the  sha- 
dow of  the  given  line  upon  that  second  surface. 

Thus :  where  the  shadow  of  a  staff,  upon  the  ground,  meets 
the  intersection  of  the  ground  with  a  house,  is  where  the  shadow 
of  the  staff  upon  the  house  begins. 


PROBLEM  XV. 

To  construct  the  shadow  of  the  upper  base  of  a  vertical  right  cone, 
upon  the  lower  nappe  of  tiie  same  cone. 

Principles. — According  to  (41)  the  upper  base  is  the  line  cast- 
ing the  shadow,  the  lower  nappe  is  the  single  curved  surface 
receiving  the  shadow,  and  any  secant  plane  containing  the  ray 
through  the  vertex,  and  cutting  the  base,  is  a  secant  plane  of 
rays  containing  two  elements  of  the  cone.  The  intersection  of 
each  of  these  elements  with  the  upper  base,  is,  in  general,  a  point 
casting  a  shadow  on  the  other  element. 

Construction.— PI.  V.,  Fig.  18.  AGL  and  A'H'— V'— A"H" 
are  the  projections  of  the  cone,  and  VN — V'E',  of  the  ray 
through  the  vertex,  which  pierces  the  horizontal  plane  at  N,E', 
and  the  plane  of  the  upper  base  at  D'",D".  As  all  the  planes  of 
rays  are  to  contain  this  ray,  their  traces  on  the  plane  of  the  lower 
base  will  all  pass  through  N,  and  those  on  the  upper  base, 
through  D'",  as  seen  in  horizontal  projection. 


SHADES  AND  SHADOWS.  33 

To  find  the  points  of  shadow,  which  fall  on  the  circumference  of 
the  lower  base  of  the  cone.  These  points  are  found  by  the  special 
method  of  (42).  The  ray  VJ— V"J',  through  the  centre  of  the 
upper  base,  pierces  the  horizontal  plane  (the  plane  of  the  lower 
base)  at  J'.  Then  a  circle  with  J  as  a  centre,  and  radius  JL 
equal  to  V"H",  that  of  the  upper  base,  will  be  the  shadow  of  the 
upper  base  on  the  plane  of  the  lower  base.  LG  is  an  arc  of  this 
shadow,  and  L,L'  and  G,G',  its  intersections  with  the  lower  base, 
are  the  required  points  of  shadow  on  the  circumference  of  that 
base.  Other  points  might  be  similarly  found. 

Ifemark. — It  will  be  observed  that,  by  this  method,  we  neces- 
sarily find,  neither  the  elements  nor  the  rays,  containing  these 
points  of  shadow  ;  nor,  consequently,  the  points  casting  them. 
To  find  the  elements,  connect  L,L'  and  G,G'  with  the  vertex 
V,V.  To  find  the  rays,  draw  the  projections  of  rays  through 
these  points  and  note  their  intersections  with  the  corresponding 
projections  of  the  upper  base.  The  latter  intersections  will  be 
those  points  of  the  upper  base,  whose  shadows  are  L,L'  and  G,G'. 

To  find  other  points  of  the  required  shadow.  These  are  found  by 
the  general  method  of  (41).  Let  A  1ST  I  be  the  horizontal  trace  of 
any  secant  plane  of  rays.  It  cuts  from  the  cone  the  two  ele- 
ments IYB,  and  AH— A'V'H".  The  point  B,B"  of  the  former, 
in  the  upper  base,  casts  a  shadow  on  the  latter  at  nf,n,  found  by 
drawing  the  vertical  projection  B' V  of  a  ray,  and  projecting  its 
intersection,  nf,  with  A'V,  at  n,  on  AY,  which  last  point  may 
also  be  found  as  below. 

Again :  let  AF  be  the  horizontal  projection  of  the  trace  of  a 
secant  plane  of  rays,  upon  the  plane  of  the  upper  base.  This 
plane  cuts  the  upper  base  at  A,A",  and  the  cone  in  the  element 
FVM — H'V  receiving  the  shadow  of  A,  A",  at  a,a'.  This  point 
was  found  by  drawing  the  horizontal  projection,  Aa,  of  a  ray, 
and  projecting  a  into  M'V  at  a'.  The  point,  a,  might,  instead, 
have  been  projected  into  the  vertical  projection,  A'V,  of  the 
same  ray.  It  might  also  have  been  found  as  n,ri  was.  (See  a,a' 
in  PI.  IV.,  Fig.  17.) 

"When  the  secant  plane  of  rays,  as  at  EK,  is  perpendicular  to 
the  vertical  plane  of  projection,  the  vertical  projections  of  the 
elements,  and  rays,  contained  in  it,  are  confounded  together  in 
its  vertical  trace  ED".  Hence  we  have  not,  in  this  case,  the 
choice  of  methods  of  procedure,  given  above,  but  must  find, 
first,  the  horizontal  projection  of  the  point  of  shadow  contained 

3 


84  GENERAL   PROBLEMS. 

in  this  plane.  Thus :  the  plane  ENK  cuts  from  the  cone  the 
elements  EV  and  KVD  ;  the  latter  of  which  determines  the 
point  D,D"  on  the  upper  base,  which  casts  a  shadow  on  the 
former  at  b,b' ;  found  only  by  drawing  the  ray  D&,  and  projecting 
b  into  the  vertical  projection,  E'V,  of  the  element  EV,  at  b'. 

The  highest  point  of  the  shadow,  is  in  the  vertical  meridian 
plane  of  rays,  CVJ.  The  point  0,0",  cut  from  the  upper  base 
by  this  plane,  casts  a  shadow  on  the  element  CY— C'V,  in  the 
same  plane,  at  e,e'. 

Through  the  points  now  found,  the  projections  of  the  curve  of 
shadow  can  be  sketched.  In  the  horizontal  projection,  the  upper 
nappe  conceals  the  shadow.  In  the  vertical  projection,  the  whole 
of  the  front  of  the  upper  nappe,  and  all  of  the  lower  nappe,  above 
the  line  of  shadow  rib'Gr,  are  shaded,  being  visible  portions  of 
the  darkened  surface  of  the  cone. 

Remark. — If  the  light  had  made  a  less  angle  with  the  hori- 
zontal plane,  than  that  made  by  the  elements  of  the  cone,  there 
would  have  been  elements  of  shade  on  the  cone  and  a  shadow 
cast  by  the  upper  circle  on  the  interior  surface.  It  is  recom- 
mended that  the  problem  should  be  solved  under  these  con- 
ditions. 

43.  In  the  following  problem  occurs  an  illustration  of  the 
special  method  of  (40),  of  the  special  method  of  (42),  and  of  a  new 
special  method,  which  may  be  called  the  method  by  one  indefinite 
rectilinear  projection  of  the  shadow,  known  by  inspection.  Thus  the 
shadow  of  a  vertical  straight  line,  upon  any  surface,  will  be 
straight,  as  seen  in  horizontal  projection.  The  method  itself,  as 
applied  in  finding  any  other  projection  of  the  shadow,  is  this. 
Having  tfie  intersection  of  the  given  projection  of  the  shadow  with  any 
element  of  the  surface  on  which  it  falls,  project  that  point  upon  OIA 
oilier  projections,  one  or  more  of  the  same  element. 

PROBLEM  XVI. 

To  construct  Hie  shadow  cast  by  the  vertical  cylinder  (Prob.  XL, 
Fifth]  upon  the  horizontal  cylinder,  and  by  the  horizontal  cylinder 
upon  tfte  vertical  cylinder. 

Principles. — The  lines  of  each  cylinder,  which  cast  the  sha- 
dows, are  their  elements  of  shade,  and  the  portions  of  the  cir- 
cumferences of  their  bases,  which  are  curves  of  shade,  as  found 


SHADES  AND  SHADOWS.  35 

iu  (Prob.  XL,  Fifth).  The  shadow  of  the  horizontal  cylinder 
upon  the  vertical  one,  will  be  found  by  the  special  method  of 
(40).  The  shadow  of  the  upper  base  of  the  vertical  cylinder  on 
the  horizontal  cylinder  will  be  found  by  the  special  method  of 
(42),  and  the  shadow  of  the  element  of  shade  of  the  vertical 
cylinder,  upon  the  horizontal  cylinder,  will  be  found  by  the  spe- 
cial method  of  (43). 

Constructions.—l0.  PI.  V.  Fig.  19.  To  find  the  shadow  of  the 
lower  front  element  of  shade,  eH — e'H,  of  the  horizontal  cylinder, 
upon  the  vertical  cylinder.  This  shadow  begins  at  t,t',  where  this 
element  of  shade  pierces  the  latter  cylinder.  Through  any  point, 
X,X',  of  the  element  of  shade,  pass  a  ray  Xx— XV.  This  ray 
meets  the  vertical  cylinder  in  a  point,  whose  horizontal  projec- 
tion is  x,  and  whose  vertical  projection  is  x ',  the  intersection  of 
the  perpendicular,  x — x',  to  the  ground  line,  with  the  vertical  pro- 
jection, XV,  of  the  ray.  In  the  same  manner,  s,s',  the  shadow 
of  the  extremity  of  the  element  of  shade,  is  found  ;  also,  r,r',  the 
last  visible  point  of  shadow  on  the  vertical  cylinder,  which  is 
cast  by  the  point,  q,q',  of  the  left  hand  base,  found  by  drawing 
the  ray,  rq  —  r'q',  backwards  from  r,  the  horizontal  projection  of 
the  extreme  left  hand  element  of  the  vertical  cylinder.  Through 
the  points  now  found,  t'x's',  the  line  of  shadow,  cast  by  et—e't', 
may  be  sketched ;  also,  #V,  the  shadow  of  the  portion,  eq—e'q', 
of  the  base,  just  mentioned. 

2°.  To  find  the  shadow  of  the  upper  base  of  the  vertical  cylinder, 
upon  the  horizontal  cylinder.  Assume  any  horizontal  plane,  as 
the  one  whose  vertical  trace  is  W,  and  containing  elements  of 
the  horizontal  cylinder,  in  which  it  is  supposed  that  points  of  the 
required  shadow  will  fall.  This  plane  cuts  two  elements  from 
the  latter  cylinder,  the  hindmost  of  which  is  projected  at  W— 
V".  By  making  a  counter-revolution,  W — v"p,  of  V",  to 
v',p,  we  find  ^Y,  the  horizontal  projection  of  this  element.  O'D' 
— OD  is  the  ray  through  the  centre  of  the  upper  base,  and  it 
pierces  the  plane  W  at  D',D.  A  circle  described  with  D  as  a 
centre,  and  radius  equal  to  OT,  is  the  shadow  of  the  upper  base 
on  the  auxiliary  plane  VV ;  then  d,  the  intersection  of  this 
shadow  with  the  element,  pV,  cut  from  the  cylinder  by  that 
plane,  is  the  horizontal  projection  of  a  point  of  the  shadow  of  the 
upper  base  of  the  vertical  cylinder  upon  the  horizontal  cylinder. 
By  projecting  d,  at  d\  into  the  vertical  projection,  W,  of  the 
element  in  which  it  lies,  we  have  both  projections  of  this  point 


36  GENERAL  PROBLEMS. 

of  shadow.  The  point  Z>,&'  is  found  in  the  same  manner  on  the 
element  nib— B'Z/  contained  in  the  auxiliary  horizontal  plane 
whose  vertical  trace  is  B' — b'.  In  like  manner  the  point  c,c',  on 
the  highest  element,  wF— E'F',  is  found  ;  the  shadow  of  the  cen- 
tre of  the  base  OT,  on  the  horizontal  plane  containing  the  high- 
est element,  being  at  C',C. 

3°.  To  find  the  shadow  of  the  front  element  of  shade,  A — A 'A", 
of  the  vertical  cylinder,  upon  the  horizontal  cylinder.  By  (43)  we 
have,  at  once,  the  straight  line  AK,  coinciding  with  the  horizontal 
trace  of  a  vertical  plane  of  rays  through  A — A' A",  for  the  hori- 
zontal projection  of  this  shadow.  It  begins  at  A,A',  where  the 
element  of  shade,  A — A' A",  pierces  the  upper  half  of  the  hori- 
zontal cylinder.  Any  other  point,  as  g',  of  the  vertical  projec- 
tion, is  thus  found.  Assume  any  element,  as  ^'J',  whose  projec- 
tion on  the  end  elevation  is  g".  By  a  counter-revolution,  the 
element  through  g"  is  found  in  horizontal  projection  at  Jg. 
Project  g,  its  intersection  with  the  shadow,  AK,  into  the  vertical 
projection,  J'g',  of  the  assumed  element,  and  g'  will  be  determined 
as  required.  Likewise  a,a',  on  the  highest  element,  and  //',  on 
the  back  element  of  shade,  produced,  are  found.  The  curve 
h'a'f,  which,  being  an  oblique  plane  section  of  a  cylinder,  is  an 
ellipse,  can  next  be  sketched  through  the  points  now  found.  To 
find  how  far  this  shadow  is  real,  draw  the  vertical  projection, 
A'V,  of  the  ray  through  the  highest  point  A,  A",  of  the  element 
of  shade  A — A' A",  note  its  intersection,  o',  with  the  indefinite 
shadow  h'a'f,  and  project  o'  at  o.  Then  Ao — h'g'o'  is  the  definite 
shadow  of  A — h'A."  upon  the  horizontal  cylinder.  At  0,0' 
begins  the  shadow,  ocbd — o'c'Vd',  of  the  upper  base,  which  is  lost 
in  the  shade  of  the  cylinder  at  d,d'. 

Remarks. — a.  When,  as  in  this  case,  the  front  element  of  shade 
of  the  vertical  cylinder  lies  to  the  left  of  the  point,  I/,  a  minute 
shadow  will  be  cast  by  it  on  the  lower  half  of  the  horizontal 
cylinder  near  I/.  When,  however,  the  same  element  of  shade 
lies  to  the  right  of  I/,  a  small  portion  of  T" — I/IF  will  cast  a 
small  shadow  on  the  vertical  cylinder. 

b.  By  constructing  the  base  G'F  in  end  elevation,  which  would 
be  a  straight  line  equal  to  GT,  and  in  GT  produced,  and  by 
constructing,  in  the  same  elevation,  the  element  of  shade 
A — A' A",  both  of  the  shadows  on  the  horizontal  cylinder  could 
have  been  found  by  the  simple  special  method  of  (40),  applied 
very  nearly  as  in  Prob.  XIII. 


SHADES  AND  SHADOWS.  37 


PfiOBLEM   XVII. 

To  find  the  axes  of  an  elliptical  shadow,  two  of  whose  conjugate 
diameters  are  known. 

Principles. — PI.  VI.,  Fig.  66.  It  has  already  been  shown 
(Elemen.  Proj.  Drawing,  Div.  IV.,  Chap.  V.)  and  (Descrip.  Geom.) 
that,  if  the  projecting  lines  of  an  object  are  oblique  to  a  plane  of 
projection,  surfaces  parallel  to  that  plane  will  be  shown,  still,  in 
both  their  true  form  and  size ;  and  that  lines  perpendicular  to 
that  plane  will  be  shown  as  parallel  lines,  whose  direction  will 
depend  on  that  of  the  oblique  projecting  lines. 

In  case  the  projecting  lines  make  an  angle  of  45°  with  the 
plane  of  projection,  as  in  "  Cavalier  Perspective "  or  Cabinet 
Projection,  the  projection  of  a  perpendicular  to  the  plane  of  pro- 
jection will  be  equal  to  the  line  itself;  since  the  line,  its  projec- 
tion, and  its  projecting  line,  taken  together,  will  form  an  isos- 
celes right-angled  triangle.  But  if  the  projecting  lines  make  any 
other  angle  than  45°  with  the  plane  of  projection,  the  projection 
of  the  line  will  be  longer  or  shorter  than  the  line  itself.  The 
latter  case  is  shown  in  Fig.  66,  which  is  a  general  oblique  pro- 
jection of  a  cube;  FG,  etc.,  being  less  than  EF,  the  edge  parallel 
to  the  plane  of  projection. 

To  view  this  figure,  now,  with  reference  to  the  present  problem, 
the  ellipse  mQnp  is  the  oblique  projection  of  the  circle  inscribed 
in  that  face  of  the  cube  which  is  perpendicular  to  its  front  face, 
EFLR,  in  the  line  LR.  Hence  the  figure  also  truly  represents, 
in  a  pictorial  manner,  the  projection  of  the  circle  MPNQ,  con- 
sidered as  vertical,  upon  the  horizontal  plane  LRHK  ;  and,  more- 
over, if  Oo,  which  thus  represents  an  oblique  projecting  line,  is 
regarded  as  a  ray  of  light,  then  mQnp  will  represent  the  shadow 
of  MPNQ  on  the  plane  LRHK. 

Construction. — Let  inn  and  pQ,  Fig.  67,  be  given  conjugate 
diameters  of  an  elliptical  shadow,  represented  by  oblique  projec- 
tion, as  diameters  of  the  shadow  of  a  vertical  circle,  upon  the 
horizontal  plane  which  is  represented  by  the  part  of  the  paper 
below  A'D' ;  a  parallel  to  mn,  through  Q,  as  in  Fig.  66.  Then 
we  have  ora,  equal  and  parallel  to  the  radius  of  the  original  circle 
casting  the  shadow,  and  Q,  as  the  shadow  of  the  foot  of  the  ver- 
tical diameter  of  the  same  circle.  Then  erect  QP,  perpendicular 


38  GENERAL  PROBLEMS. 

to  AT)',  and  equal  to  mn,  and  OQ=|PQ  is  the  radius  of  the 
original  circle. 

Next,  lines  that  are  parallel  in  reality,  as  ON  and  a  tangent  at 
P,  are  parallel  in  projection  ;  and  conjugate  diameters  are,  each, 
parallel  to  the  tangents  at  the  extremities  of  the  other.  Hence 
any  diameters  at  right  angles  to  each  other,  in  MPNQ,  will  cast 
shadows  which  will  be  conjugate  diameters  of  the  given  ellipse. 
We  therefore  seek  a  pair  of  such  diameters,  whose  shadows  shall 
be  conjugate  diameters  at  right  angles  to  each  other,  for  these 
will  be  the  axes  required. 

Now,  since  the  shadows  of  the  indefinite  radii  of  MPNQ  begin 
in  A'D',  these  radii,  and  their  rectangular  shadows,  must  be  in- 
scribed in  two  semicircles,  having  a  common  diameter  in  A'D'. 
Hence  00  is  a  chord  of  the  circle  A'OD'o,  thus  composed,  and 
gG,  perpendicular  to  it  at  its  middle  point,  meets  A'D'  in  the 
centre,  G,  of  this  circle.  Then  O  A'  and  OD'  are  the  rectangular 
radii,  whose  shadows,  A'o  and  D'o,  are  at  right  angles  to  each 
other,  and  are  each  parallel  to  the  tangents  at  the  extremities  of 
the  other.  Hence,  limiting  them  by  the  rays  a  A  and  c?D,  and 
making  oB=oA,  and  oC=0D,  we  have  AB  and  CD  for  the 
required  axes  of  the  elliptical  shadow  mQnp. 

Remark. — The  construction  just  given  is  often  found  among 
plane  problems  on  the  conic  sections,  as  it  can  readily  be  explained 
by  the  principles  of  plane  geometry.  But,  as  required  by  the 
spirit  of  the  present  subject,  it  is  here  explained  by  the  principles 
of  projections  and  of  shadows. 

44.  The  problem  of  the  Niche  is  a  favorite  one  in  Shades  and 
Shadows,  owing  to  the  considerable  number  of  points  of  peculiar 
interest  connected  with  it.     The  niche  is  a  familiar  concavity  in 
the  walls  of  halls,  staircases,  etc.,  and  consists  usually  of  a  verti- 
cal half  cylinder  of  revolution,  united  by  its  upper  base  with  a 
concave  spherical  quadrant.     The  problem  of  the  niche,  as  a 
problem  of  shadows,  consists  in  finding  the  shadow  of  the  edge 
of  shade  of  the  cylindrical  part  upon  that  part,  and  upon  the 
lower  base;    and  the  shadow  of  the   front   semicircle   of   the 
spherical  part,  upon  that  part,  and  upon  the  cylindrical  part. 

45.  Agreeably  to  the  classification  of  (12,  13)  it  becomes  de- 
sirable to  divide  this  problem,  and  to  retain  here  the  construction 
of  the  shadows  on  the  cylindrical  part,  only  leaving  the  shadow  on 
the  spherical  part  to  be  placed  with  shadows  on  other  double-curv- 


SHADES  AND  SHADOWS.  3(J 

ed  surfaces.  One  of  the  topics  of  special  interest,  in  this  problem, 
is  the  direct  construction  of  that  point  of  shadow  which  falls  on 
the  curve  of  the  upper  base ;  and  there  are  several  such  construc- 
tions, but  which  are  applicable  only  when  the  spherical  part  is 
included  in  the  problem.  Hence  it  becomes  quite  desirable  tc 
find  also  a  direct  construction  of  the  point  in  question  which 
shall  be  applicable  in  the  problem  as  given  below,  where  the 
spherical  part  is  not  recognised.  Such  constructions  will  be  found 
in  the  following  problem. 

PROBLEM  XVIII. 

Having  a  vertical  semi-cylinder  with  its  meridian  plane  parallel  to 
the  vertical  plane  of  projection,  it  is  required  to  find  the  shadow 
cast  on  its  base  and  visible  interior,  by  its  edge  of  shade,  and  by  a 
vertical  semicircle,  described  on  the  diameter  of  its  upper  base. 

Principles. — All  the  points  of  shadow,  save  that  on  the  circum- 
ference of  the  upper  base  (45),  are  found  by  the  special  method 
of  (40). 

•  Constructions.—  PI.  YL,  Fig.  20.  AcZB—  A'A^B'B"  is  the 
vertical  half  cylinder ;  AB — A^F'B"  is  the  vertical  semicircle, 
whose  shadow  is  to  be  found,  and  Aa — A"a"  is  a  ray  of  light. 

1°.  To  find  the  shadows  of  the  vertical  line,  A — A 'A".  Aa 
is  the  horizontal  trace  of  a  plane  of  rays  through  the  vertical  line, 
A — A  A/'.  This  plane  will,  therefore,  cut  the  cylinder  in  an 
element  whose  horizontal  projection  is  a.  Hence,  by  drawing 
a' — a",  the  vertical  projection  of  this  line,  and  the  rays,  a'b'  and 
A"a",  we  find  the  whole  rectilinear  shadow,  a' — a",  on  the  cylin- 
drical surface,  also  the  portion  A"6'  of  the  line  A' A",  which  casts 
the  shadow  a' — a".  The  shadow  of  A'b',  on  the  base  of  the 
cylinder,  is  Aa. 

2°.  To  find  points  of  shadow  cast  by  the  semicircle  AB — 
A"F'B".  Cc — G'c'  is  a  ray,  through  any  assumed  point,  CO',  of 
this  semicircle.  It  pierces  the  cylinder  at  the  point  whose  hori- 
zontal projection  is  c,  and  vertical  projection  c'  (40) ;  cc'  is  there- 
fore the  required  shadow  of  CO'.  Other  points  of  this  shadow 
are  found  in  the  same  manner. 

3°.  To  find  the  point  of  shadow  on  the  upper  base  of  t/ie  cylinder,  we 
will  first  use  the  following  special  method  of  auxiliary  oblique  projec- 
tions. The  ray  OS — C'S'  pierces  the  plane  of  the  upper  base  of  the 


40  GENERAL   PROBLEMS. 

cylinder  at  S'S.  If  then  the  semicircle,  AB— A"FB",  be  re- 
volved forward  about  AB — A"B"  as  an  axis,  till  it  coincides  with 
the  plane  of  the  upper  base,  CO'  will  appear  at  C",  and  C"S 
will  be  an  oblique  projection  of  the  raj  CS — C'S',  when  the 
projecting  lines  come  directly  forward  and  downward  at  an  angle 
of  45°  with  the  horizontal  plane.  Observing  the  limitation  of 
C"S  by  CS,  it  may  be  described  as  being  inscribed  in  a  circle, 
U — SKC",  whose  centre,  U,  is  found  at  the  intersection  of  AB 
with  uU,  the  bisecting  perpendicular  of  C"S.  Now  a  parallel  ray, 
similarly  inscribed  in  the  circle  E — B  AG",  will  evidently  be  the 
similar  representation  of  a  ray  which  contains  a  point  of  the  semi- 
circle AG"B  casting  the  shadow,  and  of  the  semicircle  Ac£B 
receiving  the  shadow.  The  point  on  Ac?B  will  be  the  point  of 
shadow  sought.  This  ray,  parallel  to  C"S,  will  be  a  homologous 
side  of  a  triangle  similar  to  SCO",  and  similarly  situated.  From 
these  things,  it  follows  that  the  two  rays  will  be  to  each  other  as 
the  radii,  UK  and  EA,  of  the  circumscribing  circles  SKC"  and 
BAG".  Constructing  this  proportion  at  qmo,  we  find  mq  for  the 
length  of  the  parallel  ray,  cZG".  Its  extremity,  d.  is  located  by 
drawing  Ed  parallel  to  US.  Projecting  d  at  d',  gives  dd'  as  the 
required  point  of  shadow  on  the  circumference  of  the  upper  base 
of  the  cylinder. 

Remark. — To  give  completeness  to  the  figure,  draw  dG"  parallel 
to  SO'7.  It  will  be  equal  to  qm.  Then  make  the  counter-revolu- 
tion of  AG"B  to  its  primitive  position,  when  G"  will  be  found  in 
its  primitive  position  at  GG'.  Then  drawing  the  projections,  Gd 
and  G'd',  of  the  ray  which  determines  dd',  we  shall  have  the  point 
GG',  whose  shadow  is  dd',  and,  in  plan,  the  complete  triangle 
G"Gd,  similar  to  C"CS. 

4°.  By  another  special  method,  that  of  auxiliary  spheres,  the 
ray  CS — C'S'  pierces  the  plane  of  the  upper  base  of  the  cylinder 
at  S'S,  as  before.  A  plane,  perpendicular  to  this  ray,  at  its  middle 
point  rr',  will  intersect  AB — A"B"  in  the  centre  of  a  sphere  of 
which  the  ray  is  a  chord.  Knowing  (Des.  Geom.  53i)  that  the 
vertical  trace  of  this  plane  will  be  perpendicular  to  C'S',  the  line 
r'T' — rT,  where  r'T'  is  perpendicular  to  C'S',  and  rT  parallel  to 
the  ground  line — is  a  line  of  the  plane,  hence  T'T  is  a  point  of 
its  trace  on  the  plane  of  the  upper  base.  This  trace  must  be  per- 
pendicular to  CS,  hence  it  is  the  line  TU.  UU'  is  then  the 
centre  of  a  sphere,  whose  radius  is  US,  and  in  which  the  ray 
CS— C'S'  is  inscribed. 


SHADES   AND  SHADOWS.  41 

To  find  the  ray  similarly  inscribed  in  a  sphere  whose  centre  ia 
E,E'  and  radius  EA. 

In  the  proportion  UK  :  EA::CC"  :  GG"  (=gGf)  we  find,  by  a 
construction  like  qmo,  the  fourth  term,  and  locate  it  as  at  #G'. 
Then  draw  the  ray  Gd — GW,  which  intersects  AdB— A"B"  a 
dd'  the  point  sought. 

Remarks. — a.  By  finding  the  shadow  on  the  cylindrical  sur- 
face produced,  as  at//"',  the  intersection  of  the  sketched  line  of 
shadow,  a"c'f,  with  A"B",  gives  d'  approximately  and  indirectly. 
Such  constructions  are  less  satisfactory  than  determinate  inter- 
sections, found  by  direct  constructions,  such  as  the  two  preced- 
ing. 

b.  The  straight  shadow,  a'— a",  and  the  curved  one,  d'c'a",  are 
tangent  to  each  other  at  a" ;  for  a — a! a,"  is  the  element  of  con- 
tact of  a  vertical  tangent  plane  at  a,  and  the  surface  of  the  given 
vertical  cylinder,  and  Aa—A."a"  is  the  element  of  tan  gency  of  the 
vertical  plane  of  rays,  Aa,  with  the  semi-cylinder  of  rays  having 
the  semicircle,  AB— A^F'B",  for  its  base.  Hence  a— a' a"  is  the 
intersection  of  the  tangent  planes  to  the  two  cylinders,  and  is 
therefore  also  the  tangent  line  to  their  curve  of  intersection  ;  viz. 
to  the  shadow  a"c'd'f.  (D.  G.  170.) 


SECTION  H. 

SHADES  AND  SHADOWS  ON  WAKPED  SUKFACES. 

§  I. — Shades. 

46.  The  line  of  shade  on  a  warped  surface  will,  in  general,  be 
a  curve  of  some  kind.     Its  points  must,  therefore,  be  separately 
found.     But  any  point  of  a  curve  of  shade  may  be  regarded, 
either  as  the  point  of  contact  of  a  tangent  plane  of  rays,  or  as  that 
of  a  single  tangent  ray.     But,  again,  since  we  naturally  associate 
tangencies  of  surfaces  to  surfaces,  and  of  lines  to  lines,  a  tangent 
line  to  a  given  surface  at  any  point,  is  generally  constructed  a?  a 
tangent  to  some  curve,  lying  on  that  surface  and  passing  through 
the  point. 

47.  Here  it  is  to  be  remembered  that,  as  the  consecutive  ele- 
ments of  a  warped  surface  are  not  in  the  same  plane,  a  plane,  K, 
passed  through  any  one  of  them,  E,  will,  in  general,  intersect  all 


42  GENERAL   PROBLEMS. 

the  others  on  each  side  of  it,  forming  a  curve,  C,  whose  inter 
section  with  the  element,  E,  is,  as  explained  in  Descr.  Geom. 
(212,  342),  the  point  of  contact  of  the  plane  K,  which  is  thus  a 
tangent  plane  as  well  as  a  secant  plane. 

48.  Hence  we  have  two  direct  general  methods  (22)  for  deter- 
mining any  point  of  the  curve  of  shade  on  a  warped  surface. 
One  of  these  methods  will  be  given  here,  the  other  presently. 

First  Method. —  Construct  the  point  of  tangency  of  any  plane  of 
rays,  tangent  to  the  warped  surface,  by  passing  a  plane  of  rays 
through  any  element,  and  noting  the  point  of  intersection,  P,  of  thai 
element,  with  the  curve  of  intersection  of  the  tangent  plane  with  the 
warped  surface.  The  point  of  intersection,  P,  will  he  the  point  of 
tangency  of  the  plane  of  rays,  and  hence  a  point  of  the  required  curve 
of  shade. 

49.  When  the  warped  surface  is  of  the  second  order,  either  a 
warped  hyperboloid,  or  a  hyperbolic  paraboloid,  the  plane  of  rays 
passed  through  any  element,  intersects  the  surface  in  a  second 
element,  of  the  other  generation  (Des.  Geom.),  whose  intersection 
with  the  given  element  is  the  point  of  tangency  of  the  given 
plane. 

In  illustration  of  the  method  of  (48)  take  the  following  simple 
case. 


PROBLEM  XIX. 
To  construct  tfie  curve  of  shade  on  a  hyperbolic  paraboloid. 

Principles. — By  taking  the  planes  of  projection  as  plane  direc- 
tors, and  one  of  the  directrices  vertical,  the  construction  will  be 
simphe.  The  horizontal  traces  of  the  planes  of  rays  containing 
the  horizontal  elements  will  be  parallel  to  those  elements,  respec- 
tively; and  will  cut  the  horizontal  trace  of  the  surface  in  points 
of  the  elements  which  are  parallel  to  the  vertical  plane. 

Construction. — PI.  VI.,  Fig.  21.  For  the  sake  of  defmiteness, 
call  the  horizontal  plane  the  plane  director  of  the  first  generation, 
and  the  vertical  plane,  the  plane  director  of  the  second  generation. 
Let  the  vertical  line  A— A" A'  and  the  line  A"G— A'G,  in  the 
vertical  plane,  be  the  directrices  of  the  first  generation.  Then 
divide  these  directrices  proportionally,  according  to  the  proper- 
ties of  the  surface,  and,  for  convenience,  let  these  parts  be  equal 


SHADES  AND   SHADOWS.  43 

on  each  line.  Then  AA"— A';  AC— c"C';  AD— cTI)',  etc., 
will  be  elements  of  the  first  generation.  In  order  that  the  curve 
of  shade  shall  be  on  the  visible  face  of  the  surface,  or  real,  in 
case  of  an  opaque  solid,  formed  and  placed  as  in  the  figure,  the 
light  should  come  as  shown  at  A7  in  horizontal  projection. 
Without  constructing  the  intersection  of  the  ray  through  AA' 
with  the  horizontal  plane,  by  means  of  a  given  vertical  projec- 
tion, let  7  be  assumed  as  this  intersection.  The  line  7ft,  parallel 
to  A  A",  will  then  be  the  horizontal  trace  of  the  plane  of  rays 
through  AA".  Then  na,  drawn  from  n,  the  intersection  of  the 
trace,  7w,  with  the  trace,  AG,  of  the  paraboloid,  and  parallel  to  the 
ground  line,  represents  the  element  of  the  second  generation  con- 
tained in  the  plane  A.  — In.  Its  intersection,  a,  with  the  element 
A  A",  is  the  point  of  contact  of  the  plane  of  rays,  and  hence  a 
point  in  the  required  curve  of  shade. 

Since  the  vertical  directrix  at  A  was  equally  divided  by  the 
horizontal  elements,  the  rays  through  the  points  of  division, 
being  parallel  lines,  meet  the  horizontal  plane  in  points,  on  A7, 
which  would  divide  that  line  into  equal  parts,  and  which  are  also 
points  in  traces  of  planes  of  rays  through  those  elements  respec- 
tively. Hence  6p,  parallel  to  AB,  and  le,  parallel  to  AD,  are, 
for  example,  horizontal  traces  of  planes  of  rays  containing  those 
elements.  These  planes  also  contain,  respectively,  the  elements, 
bp  and  ed,  of  the  second  generation,  which  intersect  the  former 
elements  at  b  and  d,  two  more  points  of  the  curve  of 
shade. 

Other  points  of  shade  are  similarly  found.  The  vertical  pro- 
jection of  b  is  at  b'  ]  of  d,  at  d,  etc.  AN  is  the  horizontal  pro. 
jection  of  k'N',  the  first  element  below  the  horizontal  plane. 
IK  is  the  horizontal  projection,  parallel  to  AG,  of  the  trace  of 
the  plane  of  rays  through  AG  on  the  horizontal  plane  &'N'. 
Then  by  drawing  k#,  as  in  the  previous  constructions,  we  find  g, 
the  point  of  shade  on  AG. 

Remarks. — a.  From  the  properties  of  this  surface,  the  curve  of 
shade,  that  is,  geometrically  speaking,  the  curve  of  contact  of  a 
cylinder  with  a  hyperbolic  paraboloid,  is  a  parabola,  and  hence 
is  also  a  parabolic  curve  in  both  projections. 

b.  The  problem,  as  here  given,  shows  the  shadow,  as  on  an 
awning  or  porch  roof  in  an  angle  of  a  building,  and  of  the  form 
of  a  simple  hyperbolic  paraboloid. 

50.  Second  Method  (48). — Pass  any  secant  plane  of  rays  tiirough 


44  GENERAL   PROBLEMS. 

the  warped  surface  and  construct  its  curve  of  intersection  with  the 
surface.  Then  draw  a  ray  tangent  to  this  curve  of  intersection. 
The  point  of  contact,  thus  found,  will  be  the  point  of  contact  of  the  ray 
with  the  surface,  and  hence  a  point  of  its  required  curve  of  shade. 

Either  of  the  above  direct  and  general  methods  (48,  50)  -can  be 
easily  applied  to  any  of  the  general  warped  surfaces,  with  or  with- 
out plane  directors,  and  given  by  their  elements,  in  constructions 
which  are  too  simple  to  need  formal  illustration. 

51.  When  the  warped  surface  is  a  hyperboloid,  particularly 
when  it  is  a  hyperboloid  of  revolution,  the  indirect  special  method 
of  auxiliary  tangent  surfaces  is  applicable.  This  method  is  indirect, 
as  involving  an  intermediate  surface  whose  line  of  shade  is  more 
readily  found  than  that  of  the  given  surface  (22),  and  is  special,  as 
being  applicable  to  certain  particular  surfaces  (22).  It  may  be 
stated  as  follows : 

Make  any  circular  section,  C,  of  the  given  ivarped  surface  of  revo- 
lution, the  circle  of  contact  of  an  auxiliary  tangent  cone,  having  the 
same  axis  as  the  given  surface,  and  whose  elements  of  shade,  E  and  E', 
are  easily  found. 

The  point,  T,  at  which  the  element  of  contact,  E,  of  the  auxiliary 
cone  with  &  plane  of  rays,  intersects  the  circle,  of  contact,  C,  of  the 
cone  and  the  given  surface  of  revolution,  is  the  common  point  of 
contact  of  the  plane,  the  cone,  and  the  given  surface.  That  is,  it 
is  the  point  of  contact  of  the  plane  of  rays  and  the  given  surface, 
and  hence  is  a  point  of  the  curve  of  shade  on  the  latter. 

Hence,  the  essential  principle  of  the  above  method,  stated 
abstractly,  is  this  :  Two  surfaces,  M  and  N,  being  tangent  in  a  curve 
of  contact,  K,  the  intersection,  T,  of  the  line  of  shade  of  N  with  the 
curve  of  contact,  K,  is  a  point  of  shade,  common  both  to  M  and  £T, 
and  hence  a  point  of  the  curve  of  shade  of  M. 

jRemark. — The  method  just  explained  might  be  applied  in  con- 
structing the  curve  of  shade  of  a  warped  hyperboloid ;  but  as  it 
will  be  exhibited  in  connection  with  an  analogous  double  curved 
surface  (Prob.  XXVII.)  it  is  here  omitted. 


SHADES  AND  SHADOWS.  45 


PROBLEM  XX. 

To  find  the  curve  of  shade  on  ike  common  oblique  helicoid,  in  the 
practical  case  of  the  threads  of  a  triangular-threaded  screw. 

52.  Principles. — Under  this  head  are  here  arranged  a  few  pre 
liminary  matters,  which  will  prepare  the  way  for  attention  to  the 
immediate  object  of  the  problem. 

1°.  Description  of  the  screw.  PI.  VII.,  Fig.  22. — Let  the  circle, 
whose  radius  is  AG",  be  the  horizontal  projection  of  a  vertical 
cylinder,  called  the  cylinder,  core,  or  newel,  of  the  screw.  Let 
iy"T"T'"  be  any  isosceles  triangle,  whose  base,  w'"T'",  coincides 
with  an  element  of  the  cylinder,  and  which  lies  in  a  meridian 
plane  of  the  cylinder.  Let  this  triangle  have  a  compound 
motion  ;  of  rotation  about  the  axis,  A — A' A",  of  the  cylinder, 
and  of  translation  parallel  to  that  axis ;  and  let  each  of  these 
motions  be  uniform.  With  such  motions,  each  point  of  the 
triangle,  as  w'",  I'",  or  T'",  will  describe  a  helix  (Des.  Geom.  308) ; 
the  side  Z"'T'",  not  shown,  but  corresponding  to  w'"l"f  (dotted), 
will  generate  an  upwardly  converging  zone  of  an  oblique  heli- 
coid,  and  the  side  w'"l'"  (dotted)  will  generate  a  downwardly 
converging  zone  of  a  similar  helicoid.  The  entire  triangle  will 
generate  the  volume  called  the  thread  of  the  screw.  The  heli- 
coidal  zone  generated  by  l"T"'  is  called  the  upper  surface  of  this 
thread,  and  that  generated  by  w'"l'"  is  its  lower  surface.  The 
curve,  «/"G-'aV,  etc.,  generated  by  w'",  is  called  an  inner  helix; 
while  the  curve,  l"'b"l",  generated  by  I'",  is  called  an  outer  helix. 

All  the  inner  helices  are  horizontally  projected  in  the  circle 
G"am,  and  all  the  outer  ones  in  the  circle  sqx.  The  vertical  pro- 
jection of  a  helix  is  found  by  dividing  its  horizontal  projection 
equally,  and  projecting  the  points  of  division  upon  the  successive 
equidistant  horizontal  lines,  which  mark  its  uniform  ascending 
progress. 

The  contours  of  the  helicoidal  surfaces  are  curves,  apparently 
tangent,  as  seen  in  vertical  projection,  to  the  helices.  That  por- 
tion of  the  contour  which  bounds  a  single  zone  is,  however,  so 
slightly  curved  that  it  is  sufficiently  accurate  to  represent  it  by  a 
straight  line  tangent  to  an  outer  and  an  inner  helix,  as  at  l"w'" 
(the  full  line)  and  Z'"T'".  By  thus  drawing  all  the  visible  con- 
tours, the  projections  of  the  screw  will  be  completed.  The  dotted 


46  GENERAL   PROBLEMS. 

line,  l'"w'",  whose  horizontal  projection  is  AS",  is  an  asymptote 
to  the  contour  of  the  helicoid,  being  an  element  of  the  cone  direc- 
tor (Des.  Geom.  333-4th)  whose  axis  is  A— A' A",  so  that  any 
meridian  plane  cuts  elements  from  it  and  from  the  helicoid, 
which  are  parallel. 

Tlie  core  of  the  screw  is  shown  at  V""E'"  for  a  short  distance 
above  the  horizontal  plane  &"C",  which  cuts  off  the  screw.  The 
intersection  of  the  screw  with  this  plane,  is  a  spiral  of  Archi- 
medes (D.  G.  3396),  and  is  found  by  dividing  S"S'"  into  any 
number,  as  eight,  of  equal  parts,  and  the  semicircle,  S^'G^N"", 
into  the  same  number  of  equal  parts ;  then  circles,  with  A  as  a 
centre,  drawn  through  the  points  of  S^S7",  will  intersect  the 
radii  through  the  corresponding  points  of  S"G-"N""  in  points  of 
the  spiral  required. 

2°.  To  construct  any  particular  element. — First  Method. — Any 
line,  as  A«e,  is  the  horizontal  projection  of  an  element  (D.  G.  339), 
using  a,  a  moment,  to  mark  a  point  on  Ae.  Project  a  at  a',  and 
e  at  e',  then  ae—de'  will  be  the  projections  of  the  portion  of  the 
indefinite  element  Ae,  which  lies  on  the  zone  bounded  by  the 
helices  through  the  points  S",S""  and  S">'". 

Second  Method. — Produce  the  element  through  S'^S""  and 
S'",w'"  (not  shown  in  vertical  projection)  till  it  meets  the  axis, 
A — A' A",  at  a  point  which  we  will  call  2.  Then  the  element 
through  any  point,  as  e,e',  will  meet  the  axis  as  far  above  2,  as  e' 
is  vertically  above  S"".  This  follows  from  the  uniformity  of  the 
two  motions  of  the  element  (1°). 

Remark.— As  the  construction  of  a  single  point  of  the  curve 
of  shade  of  the  screw  is  somewhat  lengthy,  by  any  method,  the 
principles  of  each  of  the  constructions  now  to.be  explained,  will 
be  given  in  immediate  connection  with  the  construction  it- 
self. 

53.  Methods  ly  assumed  helices.  These  special  methods 
consist  in  finding  the  points  of  contact  of  planes  of  rays,  which 
are  tangent  at  points  on  assumed  helices. 

The  following  will  suffice  for  illustration: 

First.  The  method  by  tangent  planes  of  given  declivity. 

Principles.  First.  The  axis  of  the  screw  being  vertical,  the 
elements,  and  the  tangents,  at  all  points  of  the  same  helix,  make, 
each,  a  constant  angle  with  the  horizontal  plane.  Second.  Hence 
all  the  tangent  planes  to  the  screw,  at  points  of  the  same 


SHADES  AND  SHADOWS.  47 

helix,  make  a  constant  angle  with  the  horizontal  plane;  they 
being  determined  by  the  lines  just  mentioned.  Third.  Hence, 
again,  the  angle  between  the  element,  and  the  line  of  greatest 
declivity,  in  these  tangent  planes,  is  constant.  This  line  of  de- 
clivity is  perpendicular  to  the  horizontal  trace  of  the  plane. 
Fourth.  A  cone,  having  the  axis  of  a  helicoid  for  its  axis,  and 
whose  declivity  is  the  same  as  that  of  all  the  tangent  planes  along 
a  given  helix,  may  be  taken  as  a  cone  director  of  the  helicoid, 
since  its  elements  make  a  constant  angle  with  those  of  the  heli- 
coid. Any  plane,  tangent  to  this  cone  will  be  parallel  to  one  of 
these  tangent  planes  to  the  helicoid.  Therefore,  a  plane  of  rays 
being  made  tangent  to  this  cone,  a  parallel  plane  of  rays  may  be 
drawn,  tangent  to  the  helicoid,  and  its  point  of  contact  will  there- 
fore be  a  required  point  of  shade. 

From  these  principles  we  have  the  following  operations  :  Draw 
any  plane,  P,  tangent  to  the  screw  at  a  point  on  a  given  helix, 
and  note  its  declivity,  L,  taken  in  a  meridian  plane,  and  which  will 
give  that  of  the  cone  director.  Then,  draw  a  plane  of  rays, 
tangent  to  this  cone.  Its  declivity,  L',  will  be  parallel  to  that 
of  the  required  tangent  plane  of  rays,  and  in  the  same  meridian 
plane.  Then  the  element,  whose  angle  with  I/  equals  the  angle 
between  L  and  the  element  of  contact  of  P,  will  intersect  the 
helix  at  the  point  of  contact  of  the  plane  of  rays,  that  is,  at  a 
point  of  shade. 

Construction. — PI.  VII.,  Fig.  22.  1°.  Let  the  assumed  helix  he  the 
outer  helix,  S"eq — S""e'.  At  any  point,  0,0',  of  this  helix,  draw  the 
tangent  line,  oO,  and  the  element  as  Ao.  The  tangent  pierces 
the  horizontal  plane  of  projection  at  O,  where  00  is  equal  to  the 
arc  osS''  (Des.  Geotn.  317),  and  the  element  pierces  the  same  plane 
at  F,F';  hence  FO  is  the  horizontal  trace  of  the  tangent  plane  to 
the  upper  surface  of  the  thread  at  0,0'.  Now  AG,  perpendicular 
to  FO,  is  the  line  of  greatest  declivity  of  this  tangent  plane,  and 
GAF  represents  the  angle  between  this  line  and  the  element  of 
contact,  AF,  of  the  tangent  plane. 

Let  the  cone,  whose  axis  is  A — A'A,"  and  whose  declivity  is 
equal  to  that  of  AG,  be  the  cone  director  for  the  helicoidal  upper 
surface  of  a  thread.  AG  meets  the  axis  at  the  point  A,F"; 
then  let  AI — A'T,  having  the  same  declivity  as  AG,  be  the  gene- 
ratrix of  the  cone  director.  The  circle  with  A  as  a  centre,  and 
radius  AI,  will  be  the  base  of  this  cone.  Through  its  vertex, 
A  A",  draw  the  ray  AB — A"B',  which  pierces  the  horizontal 
plane  of  projection  at  B',B,  hence  BD  is  the  horizontal  trace  of  a 


48  GENERAL  PROBLEMS. 

plane  of  rays,  tangent  to  the  cone  director  on  AD,  which  is  also  the 
greatest  declivity  of  this  plane,  and  is  the  horizontal  projection  of 
the  line  of  greatest  declivity  of  the  parallel  plane  of  rays,  tangent 
to  the  screw.  Then  lay  off,  on  the  outer  helix,  and  from  AD,  an 
arc  15,  equal  to  u'"o,  and  the  point  b  will  be  found  as  the  point  of 
contact  of  a  plane  of  rays,  tangent  to  the  upper  surface  of  the 
thread,  at  a  point  on  the  outer  helix.  Hence  5,  and  its  vertical 
projection,  5',  are  the  projections  of  the  required  point  of  the  curve 
of  shade  on  the  upper  surface  of  the  thread  and  on  this  helix. 

BC  is  the  horizontal  trace  of  another  tangent  plane  of  rays  tc 
the  cone  director  of  the  upper  surface  of  the  thread,  and  CA  ia 
its  element  of  contact,  and  line  of  greatest  declivity.  Therefore, 
make  C""q  equal  to  w'"o,  and  q.q'  will  be  another  point  of  the 
curve  of  shade  of  the  upper  surface  of  the  thread,  and  on  the 
outer  helix. 

2°.  To  find  a  point  of  the  curve  of  shade,  on  the  upper  surface  of  a 
thread  and  on  the  inner  helix,  S'"G"a — wf"(j'a'.  To  simplify  the 
construction,  take  an  auxiliary  horizontal  plane,  p'J',  as  far 
below  rz/ra,  on  the  element  Ao,  before  used,  as  the  horizontal 
plane  of  projection  is  below  0,0'.  Then,  drawing  AO,  the  line 
riP  will  be  the  horizontal  projection  of  a  tangent  at  n,n',  which 
pierces  the  plane  p'3'  at  P.  For,  by  the  properties  of  the 
helix  and  its  tangent,  riP  =  nG"S'"  and  00  =  osS",  while 
»G"S'":<wS"::An  :  Ao 
::AP:  AO 

hence  wP:  00  ::AP:  AO, 

a  proportion  which  is  evidently  constructed  by  limiting  the 
parallel  tangents,  at  n  and  o,  by  the  straight  line  AO. 

The  element,  An',  pierces  the  plane  p'J'  at  p',p ;  hence  pP  is 
the  trace,  on  the  plane  p'3,  of  the  tangent  plane  to  the  thread  at 
njn! ;  and  A2,  perpendicular  to  pP,  is  the  line  of  greatest  declivity 
of  this  plane.  Now,  as  before,  BD  is  the  horizontal  trace  of  the 
plane  of  rays,  tangent  to  the  cone  director,  and  AD  is  the  hori- 
zontal projection  of  its  line  of  declivity,  and  of  that  of  the  re- 
quired parallel  plane,  tangent  to  the  screw.  Then  make  4a  equal 
to  Sn  (53)  and  a,a'  will  be  the  required  point  of  the  curve  of 
shade  of  the  upper  surface  of  a  thread  and  on  the  inner  helix. 
rf*  is  a  similar  point  of  the  other  curve  of  shade,  on  the  back 
part  of  the  same  zone. 

Remarks. — a.  Points  on  other  helices,  such  as  intermediate 
ones,  can  be  similarly  found. 


SHADES  AND  SHADOWS.  49 

I.  By  going  through  with  the  preceding  constructions,  as 
applied  to  the  lower  surface  of  the  thread,  it  will  be  found  that, 
in  passing  from  the  element  of  contact  to  the  line  of  declivity  of 
a  tangent  plane,  the  motion  will  be,  in  a  (rotary)  direction,  around 
A,  opposite  to  the  direction  ouf".  Hence,  DA,  the  line  of  decli- 
vity of  the  tangent  plane  of  rays  to  the  cone  director,  being  pro- 
duced to.T,  Ax  is  the  horizontal  projection  of  the  declivity  of  this 
plane,  and  of  the  parallel  plane  of  rays  tangent  to  the  lower  sur- 
face of  a  thread.  This  being  so,  we  proceed,  according  to  the 
former  statement,  to  make  xl  equal  to  1Z»,  and  ym  equal  to  4# ; 
then  lin  is  a  curve  of  shade  on  the  lower  surface  of  a  thread. 
Likewise,  hj  is  the  horizontal  projection  of  a  second  curve  of 
shade  on  the  same  surface. 

c.  Each  of  the  curves  of  shade,  now  found  in  horizontal  pro- 
jection, has  as  many  vertical  projections,  and   corresponding 
curves  in  space,  as  there  are  helicoidal  zones  in  the  assumed  por- 
tion of  the  screw.     Thus,  ab  is  vertically  projected  at  a'b' ',  and 
a"b" ;  qr,   at  q'r'  and  in  part  near  g'  and  s';  1m,  at  I'm'  and 
I'm"  ;   and  hj,  at  li'j'  and  in  part  at  &"/',  and  m"' '. 

d.  If  we  conceive  the  entire  helicoid,  of  which  the  upper  sur- 
face, as  S""w'"a'e',  of  any  thread  is  a  zone,  to  be  represented, 
we  may  proceed  as  before  to  find  other  points  of  its  curve  of 
shade.     The  point  A,D',  where  the  curve  of  shade  meets  the 
axis,  is  the  intersection  of  the  axis  with  the  element  through 
which  the  meridian  plane  of  rays,  RAK,  is  passed,  since,  as  the 
axis  and  this  element  are  the  two  lines  cut  from  the  surface  by  a 
plane  of  rays  containing  the  supposed  element,  their  intersection 
is  the  point  of  tangency  of  the  plane  of  rays  (D.  G.  342),  and  is 
therefore  a  point  of  the  curve  of  shade.     We  thus  find  baAjh — 
b'a'D'  (the  rest  of  the  vertical  projection  not  shown)  for  a  com- 
plete branch  of  the  curve  of  shade  on  the  helicoidal  surface 
bounded  by  a  helix  of  a  radius  equal  to  AS".     Likewise  ImArq 
— I'm' is  the  branch  of  the  curve  of  shade,  on  the  similar  heli- 
coid of  which  the  lower  surface  of  a  thread  is  a  zone. 

e.  The  point,  A,D',  is  also  the  point  of  tangency  of  the  plane 
of  rays  RAK,  from  the  particular  properties  of  the  helicoid,  as 
well  as  from  those  of  warped  surfaces  in  general,  as  explained  in 
the  last  remark.     For  the  axis  A — A' A"  is  the  limiting  helix, 
generated  by  that  point  of  an  element  which  is  in  the  axis ;  but, 
being  straight,  it  is  its  own  tangent,  throughout  its  whole  extent; 
hence,  A,D',  like  other  points  of  contact,  e,e',  etc.,  of  tangent 

4 


50  GENERAL    PROBLEMS. 

planes,  is  the  intersection  of  a  tangent  to  the  helix,  through  A,D', 
with  the  element  through  the  same  point. 

54.  Second.  The  method  by  helical  translation  of  a  tangent 
plane. 

Points  of  shade,  on  the  lower  surface  of  a  thread,  may  be  found 
by  the  method  of  [53-2°  (Rem.  &)].  For  the  sake,  however,  of 
illustrating  a  different  construction,  they  will  be  found  by  the 
special  method  just  named. 

Principles. — Construct  a  simple  tangent  plane,  P,  to  the 
helicoid  at  any  point  of  a  given  helix,  find  its  trace,  T,  on  any 
horizontal  plane,  K;  and  note  its  intersection,  T,  with  the  axis  of 
the  helicoid.  A  ray  through  I  will  pierce  the  plane  K  in  the 
trace,  T,  if  P  is  a  plane  of  rays.  If  this  ray  does  not  meet  the 
trace  T,  revolve  it  about  the  axis  of  the  helicoid,  when  it  will 
generate  a  right  cone,  from  which  will  be  cut  two  elements,  E  and 
E',  by  the  plane  P.  By  revolving  either  of  these  to  coincide  with 
the  ray,  and  by  revolving  the  point  of  contact  of  P  through  an 
equal  angle,  keeping  it  also  on  the  helix,  we  shall  have  the  point 
of  con  tact  of  P,  when  translated,  helically,  so  as  to  be  a  tangent 
plane  of  rays.  This  point  of  contact  will  therefore  be  a  point  of 
the  required  curve  of  shade  on  the  given  helix. 

Construction. — PI.  VII.,  Fig.  22.  Let  the  tangent  plane  be 
constructed  at  the  point  S",£'",  and  let  its  trace  be  found  on  the 
horizontal  plane  S'K'.  Thus  S"A— Z"'U,  the  element,  through 
S",lf",  of  the  helicoidal  zone,  forming  the  lower  surface  of  the 
thread  considered,  pierces  the  plane  S'K'  at  S',S.  The  plane 
S'K'  being  here  taken  at  a  distance  above  S'T"  corresponding  to 
a  quarter  revolution  of  S'V"  around  the  axis  A — A'A"  we  have 
S"T"",  equal  to  a  quadrant  of  the  circle  S"#7,  as  the  tangent  at 
S'T",  piercing  the  plane  S'K'  at  T"".  Hence  T""ST  is  the 
trace,  on  the  plane  S'K',  of  the  tangent  plane  at  S",Z'".  Now, 
as  this  tangent  plane  contains  the  element  £"'U,  it  cuts  the  axis 
at  U,  and  we  wish  next  to  find  whether  a  ray  through  U  pierces 
S'K'  in  the  trace  TT"".  For  lack  of  room  on  the  plate,  draw 
AK — UK',  the  position  of  a  ray  through  A,TJ,  after  a  revolution 
of  180°  from  its  primitive  position,  AR.  (The  angle  K'UA'  is 
equal  to  the  angle  B'A"A'.)  The  revolved  ray  pierces  the  plane 
S'K'  at  K',K,  one  point  of  the  base,  KQT,  of  the  cone  generated 
by  the  revolved  ray  AK — U'K.  As  this  base  does  not  contain 
R,  the  ray  in  its  primitive  position,  AR,  does  not  pierce  the 


SHADES  AND  SHADOWS.  51 

plane  S'K',  in  the  trace,  TT"",  of  the  tangent  plane.  Hence  this 
tangent  plane  is  not  a  plane  of  rays.  But  AT  and  AQ  are 
elements  cut  from  the  cone,  just  described,  by  the  tangent  plane, 
and  may  be  considered  as  two  positions  of  the  ray  AR,  after 
being  revolved  about  the  axis,  A — A'A",  till  it  becomes  a  line 
of  the  tangent  plane.  Therefore,  the  ray  being  at  T  A,  when  the 
point  of  contact  of  the  tangent  plane  is  at  S'T",  make  S"h=7  — 5, 
and  project  h  at  h',  then  will  hji'  be  the  point  of  contact  of  the 
tangent  plane  after  being  revolved,  with  an  ascending  helical 
motion  about  A — A'A",  till  by  the  return  of  TA  to  being  a  ray 
AR,  it  becomes  a  tangent  plane  of  rays.  Likewise,  AQ  being 
the  revolved  position  of  the  ray,  RA,  when  Si", I'"  is  the  point  of 
contact  of  the  tangent  plane,  make  S"l=S — 5,  and  project  I  at  I', 
and  1,1'  will  be  another  point  of  contact  of  the  tangent  plane, 
after  moving,  with  a  descending  helical  motion,  till  it  becomes, 
again,  a  tangent  plane  of  rays.  Hence,  finally,  h,ti  and  1,1'  are 
points  of  the  two  curves  of  shade  on  the  lower  surface  of  a  thread 
of  the  screw. 

In  a  similar  manner  j,f  and  m,m'  could  have  been  found. 
hj — h'j',  being  in  front  of  the  meridian  plane  SI,  is  visible  in 
vertical  projection.  Im — I'm'  is  wholly  invisible. 

55.  A  little  consideration  of  the  properties  of  the  helicoid,- 
explained  in  Descriptive  Geometry  (Des.  Geom.  339),  will  enable 
the  student  to  understand  the  following  additional  properties, 
which  are  here  applied  in  another  construction  of  the  required 
curve  of  shade,  by  the  special  method,  called  the  method  by 
assumed  elements. 

Principles. — The  development  of  a  given  portion  of  any  helix 
— its  axis  being  conceived  to  be  vertical — is  the  hypothenuse  of 
a  right-angled  triangle,  whose  altitude  is  the  vertical  height 
between  the  extremities  of  the  given  portion,  and  whose  base  is 
the  development  of  the  circular  arc  forming  the  horizontal  pro- 
jection of  the  same  portion. 

56.  Portions  of  different  helices  on  the  same  helicoid,  but 
included  between  the  same  meridian  planes,  may  be  called  homo- 
logous.    Points  in  which  they  intersect  the  same  element,  may  be 
called  homologous,  or  corresponding  points. 

Since  all  points  of  the  generatrix  of  a  helicoid  ascend  at  the 
same  rate,  the  homologous  portions  of  helices  just  mentioned, 
will  be  of  equal  height,  between  their  extremities  ;  that  is,  they 


52  GENERAL    PROBLEMS. 

•will  be  included  between  pairs  of  equidistant  horizontal  planes. 
Tangents  to  these  helices  at  corresponding  points,  and  limited  by 
the  horizontal  planes  which  include  these  helices,  may  be  called 
homologous  tangents.  Their  horizontal  projections  will  be  par- 
allel, and  equal  to  the  concentric  circular  arcs  which  are  the 
horizontal  projections  of  the  respective  helical  arcs  to  which  they 
are  tangent.  But  these  arcs,  and  hence  their  tangents,  are  pro- 
portional to  their  radii ;  hence  the  horizontal  projections  of  these 
tangents  will  terminate  in  a  straight  line  through  the  horizontal 
projection  of  the  axis  (53-2°),  which,  with  the  tangents,  and 
their  respective  radii,  will  form  similar  triangles. 

56.  Once  more,  equidistant  horizontal  planes  will  intercept 
equal  portions  of  any  element ;  hence,  if  two  such  planes  contain, 
respectively,  the  point  of  contact  and  the  trace  of  a  tangent 
plane  through  any  element ;  and  if  another  pair  of  such  planes 
contain  the  point  of  contact  and  the  trace  of  another  tangent 
plane  through  the  same  element,  the  distance  between  the  trace 
and  the  point  of  contact  of  one  tangent  plane  will  be  equal  to  the 
distance  between  the  point  of  contact  and  the  trace  of  the  other 
tangent  plane,  these  distances  being  measured  on  the  element 
contained  in  both  of  these  planes.  Eecollecting,  too,  that  the 
helix  coincides  with  its  tangent  when  developed,  we  have  the 
following. 

Construction. — PI.  VII.,  Fig.  22.  Assume  any  element,  Ae, 
between  the  previously  found  points,  a  and  5,  on  the  helices. 
Making  the  tangent,  eL,  equal  to  the  arc  esS" ;  the  point  L 
is  the  intersection  of  this  tangent  with  the  horizontal  plane  of 
projection  (D.  G.  317).  Then  H'H  being  the  intersection  of  the 
element  Ae — EV  (52-2°)  with  the  horizontal  plane,  HL,  is  the 
horizontal  trace.of  the  tangent  plane  to  the  screw,  at  ee'  (D.G.  347.) 

Next,  by  drawing  a  ray  (not .  shown  in  vertical  projection) 
through  A,E',  it  will  pierce  the  horizontal  plane  of  projection  at 
6  in  the  line  AB,  and  6HM  will  be  the  horizontal  trace  of  a 
plane  of  rays  through  the  assumed  element. 

Now,  if  the  trace,  HM,  of  this  plane  of  rays,  were  found  on  a 
horizontal  plane  as  far  below  its  required  point  of  contact,  as  the 
trace,  II L,  is  now  below  ee',  this  new  trace  would  be  parallel  to 
HM,  the  tangent  at  its  point  of  contact  would  be  parallel  to  eM, 
and  their  distance  apart,  measured  on  Ae,  would  be  equal  to  ell. 
Hence  the  portion  of  the  new  tangent,  cut  off  by  the  new  trace, 
would  be  equal  to  eM.  But  all  tangents  to  helices  at  points  on 


SHADES  AND  SHADOWS.  63 

Ae,  and  included  between  pairs  of  equidistant  horizontal  planes, 
will  terminate  in  AL,  as  seen  in  horizontal  projection  (53-2°)  ; 
hence,  transfer  M  to  N,  on  AL,  by  a  line  MN  parallel  to  HA, 
draw  Nc,  parallel  and  equal  to  dVl,  and  c,c'  will  be  the  point  of 
contact  of  the  plane  of  rays,  HM,  with  the  screw,  and  on  the 
assumed  element  Ae.  Hence  c,c'  is  a  point  of  the  required  curve 
of  shade. 

Remark. — By  the  above  method,  as  well  as  by  that  of  (53-2°), 
we  can  find  the  indefinite  curve  of  shade  on  a  complete  helicoid. 

Discussion. 


FIRST.  The  meridian  plane,  RAK,  which  is  tangent  to  the  heli- 
coid at  A,D',  makes  an  angle  of  90°  with  the  horizontal  plane. 
At  a  point,  analogous  to  any  point  as  e.  but  on  the  helix  which 
is  at  an  infinite  distance  from  the  axis,  the  tangent  eL  would  be 
infinite,  hence  HL  would  then  be  parallel  to  eL ;  that  is,  AH 
would  be  both  the  line  of  declivity  and  the  element  containing 
the  point  of  contact  of  the  tangent  plane.  Hence  when  the 
element  of  contact  of  a  plane,  tangent  to  a  helicoid,  is  also  the 
line  of  declivity  of  that  plane,  the^om^  of  contact  is  on  a  helix 
at  an  infinite  distance  from  the  axis. 

SECOND.  To  find  tlie  helix,  II,  the  tangent  planes  for  all  points  of 
which  make  a  given  angle,  /3,  with  the  horizontal  plane ;  /3  being 
greater  than  oc.  Construct  a  cone,  having  the  axis  A — A'A"  for 
its  axis,  and  its  vertex  at  the  intersection  of  an  element,  E,  of  the 
helicoid,  with  the  axis,  and  let  all  its  elements  make  an  angle 
equal  to  /3  with  the  horizontal  plane  of  projection.  Then  con- 
struct a  plane,  through  the  element  E,  and  tangent  to  this  cone. 
This  plane  will  make  an  angle  p  with  the  horizontal  plane,  and 
its  point  of  contact,  on  E,  found  as  in  (57  Const.),  will  be  a  point, 
p,  of  the  required  helix.  Having  the  point  p,  the  helix  can 
readily  be  constructed. 

THIRD.  Now  let  /3  be  the  angle  made  by  the  rays  of  light  with 
the  horizontal  plane  of  projection,  and  oc,  as  before,  the  angle 
made  by  the  elements  of  the  helicoid  with  the  same  plane.  The 
angle  /3  may  be  greater  than,  equal  to,  or  less  than  oc.  In  PI. 
VII.,  Fig.  22,  /3  is  less  than  oc,  and  the  curves  of  shade  are  open 
curves  with  infinite  branches.  If  ft  were  equal  to  oc,  a  tan- 
gent plane  of  rays  would  be  tangent  to  the  helicoid  at  the  inter- 


54  GENERAL  PROBLEMS. 

section  of  the  element,  AZ",  with  the  infinitely  remote  helices, 
one  on  each  nappe  of  the  helicoid.  At  the  lower  of  these  points, 
b  and  q  would  unite;  and  at  the  upper  one,  I  and  h  would  unite. 
When  /3  is  greater  than  oc,  find  the  helix,  H,  described  above, 
and  apply  to  it  the  method  of  (53-2°),  but  giving  to  the  elements 
of  the  auxiliary  cone  an  angle  of  declivity  equal  to  /3.  It  will 
then  be  found,  that  but  one  plane  of  rays  can  be  drawn,  tangent 
on  the  helix  H,  and  that  the  points  q  and  b  will  unite  on  that 
helix.  Also  I  and  h  will  similarly  unite  on  a  similar  helix  of 
the  upper  nappe,  and  the  curves  of  shade  will  be  closed  curves, 
wholly  within  the  helix  H. 

FOURTH.  If  we  substitute  for  the  triangular  generatrix  of  the 
thread  of  the  screw,  a  square,  having  one  of  its  sides  in  an  ele- 
ment of  the  cylindrical  core  of  the  screw  (52),  a  square-threaded 
screw  will  be  formed,  the  upper  and  lower  surfaces  of  whose 
threads — the  axis  being  vertical — will  be  right  helicoids,  having 
the  horizontal  plane  for  their  plane  director.  For  all  practical 
cases  the  angle  0  is  not  0°,  hence  for  the  case  of  the  curve  of 
shade  on  the  square  threaded  screw,  /3  is  greater  than  oc,  and 
the  curve  of  shade  would  therefore  be  closed.  Usually,  also,  £, 
and  the  rate  of  ascension  of  the  threads,  have  such  values  that 
the  curve  of  shade  would  be  wholly  within  the  core ;  hence, 
practically,  there  is  no  problem  of  the  curve  of  shade  on  the  heli- 
coidal  surfaces  of  the  square  threaded  screw. 

The  elements  of  shade  on  its  cylindrical  surfaces  are  readily 
found. 

58.  Either  of  the  direct  general  methods  (48,  50)  may  be  applied 
in  the  construction  of  curves  of  shade  on  warped  surfaces,  but 
owing  to  the  ease  with  which  a  plane  is  drawn  tangent  to  a 
warped  surface  of  the  second  order,  the  following  indirect  general 
met/wd  will  be  more  convenient. 

Observing  that  a  hyperbolic  paraboloid  may  be  more  easily 
represented  in  any  position  than  a  warped  hyperboloid,  make 
any  element  of  the  given  warped  surface  the  element  of  contact  of  an 
auxiliary  tangent  hyperbolic  paraboloid-  The  point  of  contact  of  a 
tangent  plane  through  this  element,  can  easily  be  found,  and  it  will  be 
the  point  of  contact  on  both  surfaces.  (Des.  Geom.  343.) 

In  illustration  of  this  method  the  following  problem  is  given. 


SHADES  AND  SHADOWS  55 


PROBLEM  XXL 

To  construct  points  of  the  curve  of  shade  of  a  conoid. 

Principles. — Let  the  curved  directrix  of  the  conoid  lie  in  the 
horizontal  plane,  let  its  rectilinear  directrix  be  perpendiculai  to 
the  vertical  plane  of  projection,  and  let  this  plane  be  taken  as  the 
plane  director  of  the  conoid,  and  of  the  elements  of  the  first 
generation  of  the  auxiliary  paraboloid.  Let  the  directrices  of  the 
first  generation  of  the  paraboloid  be  the  line  of  striction  of  the 
conoid,  and  a  tangent  to  its  base  at  the  foot  of  the  assumed  ele- 
ment of  the  conoid,  through  which  a  plane  of  rays  shall  be  drawn. 
The  horizontal  plane  of  projectiQii  will  then  be  th.e  plane  director 
of  the  elements  of  the  second  generation  of  the  paraboloid. 

Construction. — PI.  VIII.,  Fig.  24.  MEK  is  the  curved,  and 
A  A" — A'  is  the  rectilinear  directrix  of  the  conoid.  Let  TB—  T'E' 
be  the  element  of  the  conoid,  on  which  it  is  proposed  to  find  a 
point  of  the  curve  of  shade.  Through  any  point,  as  A",  A',  of  this 
element,  draw  the  ray  A"R — A'R'.  This  ray  pierces  the  hori- 
zontal plane  of  projection  at  R,  and  the  assumed  element  pierces 
it  at  E,E',  hence  ERG  is  the  horizontal  trace  of  the  plane  of  rays 
through  TE— T'E'.  GA',  its  vertical  trace,  is  parallel  to  E'T',  since 
that  element  is  parallel  to  the  vertical  plane  of  projection. 

Now,  making  AF  tangent  to  MEK  at  E,  it  follows  that 
AA" — A'  and  AF  are  the  directrices  of  the  elements  of  the  first 
generation  of  the  auxiliary  paraboloid,  which  is  tangent  to  the 
conoid  along  the  element  ET — E'T'. 

F/i — FA'A'  is  another  element  of  the  paraboloid,  and  it  is  cut  by 
the  plane  of  rays,  through  ET — E'T',  at  A 'A,  which  is,  therefore, 
a  point  in  an  element  of  the  second  generation.  But  the  hori- 
zontal plane  of  projection  is  the  plane  director  of  such  elements, 
AF  being  one  of  them ;  hence  A'T'  is  the  vertical  projection  of 
the  element,  and  T'  the  vertical  projection  of  its  intersection  with 
ET— E'T'.  Projecting  T'  at  T,  we  have  T,T'  as  the  intersection 
of  elements,  one  of  each  generation,  of  the  paraboloid.  These  ele- 
ments, being  contained  in  the  same  plane,  viz.  the  plane  of  rays, 
T,T'  is  the  point  of  tangency  of  this  plane  of  rays  with  the  para- 
boloid. But,  as  the  paraboloid  and  conoid  are  tangent  to  each 
other  throughout  the  element  ET— E'T',  the  point  T,T'  is  the 


56  GENERAL  PROBLEMS. 

point  of  tangency  of  the  plane  of  rays  with  the  conoid,  also ;  and 
hence  is  a  point  of  the  required  curve  of  shade  of  the  conoid. 

Remarks. — a.  If  the  direction  of  the  vertical  trace,  G'h',  of  the 
plane  of  rays,  were  unknown,  it  would  be  found  by  drawing, 
through  any  point  of  ET — E'T',  a  horizontal  line,  as  A"g — A'g', 
parallel  to  EG,  whose  intersection,  g,g',  with  the  vertical  plane, 
woulcf  be  a  point  of  G'h'. 

b.  The  line  hT,  being  a  horizontal  line  of  the  plane  of  rays,  is 
parallel  to  GE,  the  horizontal  trace  of  that  plane.   Also,  a  vertical 
line  at  A,  being  a  directrix  of  the  second  generation  of  the  parabo- 
loid, AT  necessarily  meets  FE  at  A.     Hence  to  find  T,T'  we  have, 
practically,  the  following  very  simple  construction,     first,  draw 
EH.     /Second,  through  A  draw  Ah  parallel  to  ER.     Third,  note 
T,  the  intersection  of  Ah  arid  ET,  and  project  it  at  T'  on  ET'. 

Any  other  points  of  the  cuyve  of  shade  may  be  similarly 
found. 

c.  It  follows,  from  the  constructions  now  given,  that  either 
projection  of  the  curve  of  shade  of  a  conoid  may  be  found  inde- 
pendently of  the  other. 

Discussion. 

FIRST. — Preliminary.  This  discussion  may  be  conducted  with 
reference  to  the  right  conoid  with  a  circular  base,  PL  VIII. ,  Fig. 
25,  having  the  vertical  plane  of  projection  for  its  plane  director, 
and  the  line,  AA" — A',  for  its  line  of  striction. 

Let  cR  be  the  ray  through  the  point  c,A'  of  the  element  ec. 
Then  eR  is  the  trace  of  the  plane  of  rays  through  ec.  By 
(Rern.  b)  above,  note  a,  where  the  tangent  at  e  meets  A"A'. 
Then  at,  parallel  to  eR,  meets  ec  at  <,  the  point  of  shade  in  the 
branch  Ath,  and  determined  by  the  tangent  plane  of  rays  eR. 

The  vertical  projection  of  t  will  be  on  the  vertical  projection 
of  ec,  and  in  the  upper  nappe  of  the  conoid.  By  similar  con- 
structions, four  branches  of  the  curve  of  shade  may  be  found, 
whose  horizontal  projections  are  Ath;  AFra;  A"h,  and  Af'g. 

SECOND.  In  proceeding  with  the  discussion,  the  following  cases 
will  appear. 

First.  The  horizontal  trace  of  a  plane  of  rays  through  AA" 
— A',  may  intersect  the  base  of  the  conoid. 

1st.  In  any  manner,  as  at  BE",  PI.  VIII.,  Fig.  25. 
2d.  In  the  diameter  AA"— N'. 


SHADES  AND  SHADOWS.  57 

Second.  This  trace  may  be  tangent  to  the  base  of  the  conoid. 
Third.  The  same  trace  may  be  wholly  exterior  to  the  same 
base. 

1st.  At  a  finite  distance. 
2d.  At  an  infinite  distance. 

TIIIRD.  For  the  1st  form  of  the  first  case,  it  appears:  1°.  That 
the  points  of  shade,  consecutive  with  A,  A'  and  A",  A',  being 
points  of  tangency  of  parallel  rays,  on  elements  consecutive 
with  the  vertical  elements  at  A  and  A"]  A, A7  and  A", A' 
are  cusp  points  of  the  first  species,  as  seen  in  horizontal  projec- 
tion. 

2°.  That  the  element  which  pierces  the  horizontal  plane  at  K,K' 
is,  in  space,  an  asymptote  to  the  branches  Ah,  on  the  upper,  and 
A"<7,  on  the  lower  nappe  of  the  conoid.  For,  the  tangent  at  K, 
analogous  to  ea,  will  be  parallel  to  AX/A,  and  hence  the  points  a' 
and  a",  analogous  to  a,  and  determined  by  this  tangent,  will  be 
at  infinite  distances,  on  A  A"  in  each  direction  from  J.  But 
lines  analogous  to  at,  and  through  such  points,  a', a",  and  parallel 
to  the  trace  of  the  plane  of  rays  through  JK — A'K',  would  meet 
HK — H/'K'  only  at  infinitely  distant  points  of  shade,  analogous 
to  t.  Likewise,  the  element  which  pierces  the  horizontal  plane 
at  H,H'  is  an  asymptote  to  Am  and  A"Tc. 

3°.  The  points  A,T,T"  and  A"  are  all  vertically  projected  at 
A',  hence  the  branches  Ah  and  A"g,  each  of  which  is  partly  on 
both  nappes  of  the  conoid,  form  loops,  as  seen  in  vertical  pro- 
jection ;  to  which  N'T'  and  E'K'  are  tangent,  at  the  multiple 
point  T'. 

4°.  All  those  parts  of  these  curves  of  shade  are  real,  or  would 
exist  on  the  exterior  of  a  conoidal  opaque  solid,  like  the  figure, 
from  whose  points  rays  would  pierce  the  horizontal  plane  of 
projection,  outside  of  the  base  of  the  conoid. 

For  the  2d  form  of  the  first  case,  AA"  and  A'N'  would 
be  the  projections  of  a  ray  of  light ;  and  the  loops  in  the  vertical 
projection  would  disappear. 

FOURTH.  In  the  second  case,  the  construction  of  (Fig.  24 — 
Eem.  b)  will  give  a  curve  of  one  branch,  in  which  T  and  T"  will 
unite  at  J,  and  loops  will  reappear  in  vertical  projection,  having 
A'N'  and  H"K'  for  their  tangents  at  the  point  J,T'. 

In  this  case,  the  element  JK — A'K'  will  also  be  an  element 
of  shade,  being  an  element  of  tangency  of  a  plane  of  rays,  through- 
out its  whole  extent.  On  this  element,  therefore,  the  conoid  ia 


58  GENERAL  PROBLEMS. 

said  to  be  a  developable  single  curved  surface.  Along  the  ver- 
tical elements  at  A  and  A"  planes  may  also  be  tangent. 

Remark. — FIFTH.  In  the  1st  form  of  the  third  case,  the  loops  of 
the  vertical  projection  will  again  disappear,  and  two  branches, 
as  seen  in  horizontal  projection,  will  proceed  from  A  to  the  right, 
and  two  from  A"  to  the  left. 

In  the  2d  form  of  this  case,  the  ray  of  light  will  be  horizontal. 
The  vertical  projection  will  be  of  the  same  form  as  in  the  1st 
form  of  this  case,  but  in  horizontal  projection,  the  two  branches 
from  A,  and  the  two  from  A",  will  coincide. 


§  n. — Shadows. 

59.  The  fundamental  condition  for  determining  a  plane,  is, 
that  it  shall  pass  through  three  points  taken  at  pleasure.     Equi- 
valent derived  conditions  are,  that  it  shall  contain  a  point  and  a 
straight  line;  or  shall  contain  one  straight  line,  and  shall  be  paral- 
lel to  another.    All  the  elements  of  a  cone  have  a  common  point, 
the  vertex,  hence,  as  already  seen  (41  5),  it  is  easy  to  pass  a  sys- 
tem of  planes;  each  containing  a  given  line,  as  a  ray,  and  the 
vertex;  and  these  will  be  planes  of  rays,  cutting  the  cone  in  rec- 
tilinear elements.     Likewise,  all  the  elements  of  a  cylinder  have 
a  common  direction :  hence,  as  before  seen  (41  a),  a  system  of 
planes  of  rays  can  be  readily  passed  through  the  various  elements 
of  a  cylinder.     But  let  a  plane  be  passed  through  a  given  ray, 
and  a  given  point  of  a  warped  surface  ;  only  one  element  would 
generally  pass  through  that  point,  and  that,  generally,  would  not 
happen  to  be  in  the  supposed  plane.     Again,  let  a  plane  be  passed 
through  a  given  ray  and  parallel  to  some  given  line;  only  one 
element  would  generally  be  parallel  to  that  line,  and  that  element 
would  not  commonly  happen  to  be  in  the   plane  employed. 
Hence  we  have  the  Theorem,  that,  as  the  elements  of  a  warped  sur- 
face have  neither  a  common  point,  nor  a  common  direction,  no  simply 
arranged  system  of  planes  of  rays  can  be  made  to  cut  it,  each  in  a 
right-line  element. 

60.  Hence  (41)  in  its  application  to  warped  surfaces  gives  the 
following  general  method. 

Pass  any  secant  plane  of  rays  through  any  point  lasting  a 
shadow,  and,  by  constructing  its  intersections  with  a  number  of 
elements,  find  its  curve  of  intersection  with  the  warped  surface, 


SHADES  AND  SHADOWS.  59 

Then,  the  intersection  of  a  ray  through  the  given  point,  with  the 
curve  found,  will  be  a  required  point  of  shadow. 

61.  So  much  construction  for  each  point  of  shadow,  would  be 
extremely  tedious,  hence  we  have  the  following  inverse  general 
method.     Pass  a  plane  of  rays  through  any  assumed  element  of 
the  warped  surface,  and  note  the  point,  P,  cut  by  it,  from  the 
line  casting  the  shadow ;  then,  the  ray  through  this  point  will 
pierce  the  assumed  element  in  a  point  of  shadow. 

It  is  evident,  however,  that,  unless  the  plane  of  rays  happens 
to  be  also  a  projecting  plane,  the  construction  of  P  will  generally 
be  as  tedious  as  that  of  the  curve  of  intersection,  in  (60).  Hence 
various  special  methods,  soon  to  be  explained,  are  mostly  era 
ployed  in  finding  shadows  on  warped  surfaces. 

62.  Taking  up  the  topics  mentioned  in  shades  on  warped  sur- 
faces, and  in  the  same  order,  we  have,  first,  the  shadow  of  the  upper 
base  of  a  warped  hyperboloid  upon  that  surface,  the  construction  of 
which  will  be  explained  in  a  subsequent  problem  of  double- 
curved  surfaces. 

To  find  the  highest  and  lowest  points.  Find  the  intersection  of  a 
ray,  drawn  through  the  intersection  of  the  meridian  plane  of  rays 
with  the  circumference  of  the  upper  base,  with  the  meridian  curve 
contained  in  that  plane. 

This  construction  involves  an  easy  application  of  the  general 
method  of  (60)  whenever  the  warped  hyperboloid  is  one  of  revolu- 
tion, with  its  axis  perpendicular  to  either  plane  of  projection,  for 
then  its  meridian  curve  is  readily  found. 

63.  To  find  intermediate  points.     Employ  the  special  method  of 
auxiliary  shadows  (42).     Any  horizontal  plane,  P,  will  intersect 
the  given  surface,  its  axis  being  vertical,  in  an  ellipse,  or  circle, 
C,  if  the  surface  be  one  of  revolution.     The  shadow  of  the  upper 
base  on  the  plane  P  will  be  a  curve  equal  to  that  base  (33),  and 
the  intersections  of  this  shadow  with  C,  will  be  points  of  shadow 
falling  on  the  given  surface. 

64.  In  PL  VI.,  Fig.  21,  the  shadow  of  the  curve  of  shade  of  the 
hyperbolic  paraboloid  upon  the  horizontal  plane,  is  the  parabolic 
curve  L#,  formed  by  the  intersection  of  the  traces  of  the  succes- 
sive tangent  planes  7n,  etc.     L7,  equal  and  parallel  to  aA,  is  the 
shadow  of  aA,  and  7  A  is  the  shadow  of  A — A" A'. 


60  GENERAL  PROBLEMS. 

PROBLEM  XXII. 

To  find  Hie  several  shadows  cast  by  a  triangular- threaded  screw 

upon  itself. 

65.  These  shadows  are: 
FIRST.  The  shadow  of  the  nut. 

1st.  Of  any  unknown  point  on  an  edge  of  shade  of  the  nut. 
2d.  Of  some  assumed  point  on  such  an  edge. 
SECOND.  The  shadow  of  the  curves  of  shade. 
1st.  On  an  assumed  element. 
2d  On  an  assumed  helix. 
3d.  On  another  branch  of  the  curve  of  shade. 
THIRD.  The  shadow  of  the  outer  helix. 
1st.  On  any  assumed  element. 
2d.  On  any  helix. 
3d.  On  any  particular  element. 

FOURTH.  There  may  also  be  associated  with  the  preceding,  the 
shadow  of  the  entire  screw  and  nut,  on  the  horizontal  plane  of 
projection. 

1st.  The  determination,  by  inspection,  of  all  the  lines  casting 

this  shadow. 
2d.  The  construction  of  the  shadow  of  any  assumed  point, 

which  will  cast  a  shadow  on  the  horizontal  plane. 
Each  of  the  above  topics,  including  both  its  principles  and 
construction,  so  far  as  these  need  be  separately  mentioned,  will 
now  be  taken  up  in  the  above  order. 

66.  To  find  the  shadow  of  the  nut  upon  the  screw. 
1st.  The  shadow  of  any  unknown  point  of  the  nut. 
Principles. — This  shadow  is  found  on  the  principle,  that  any  line 

L  in  a  plane  of  rays,  R,  through  an  edge,  E,  of  the  nut,  will  meet 
the  intersection  of  B  with  a  given  surface,  in  a  definite  point  of 
shadow  of  some  point  in  the  edge,  E  ;  though  the  line  be  not, 
itself,  a  ray. 

This,  which  is  an  inverse  method  (61)  may  be  called  the  method 
by  assumed  lines  in  planes  of  rays.  It  is  usually  applicable  only 
when  the  line  E  is  straight,  since  a  plane  of  rays  can  usually  be 
passed  through  a  straight  line  only  (6). 

Construction.— PL  VIII.,  Fig.  2  J.  The  edges  of  the  nut,  which 
cast  shadows,  are  the  lower  ones,  projected  at  JK  and  KL,  and 
portions  of  JI  and  LN",  all  of  which  are  vertically  projected  in 


SHADES  AND  SHADOWS.  61 

the  line  J'L'.  Beginning  with  IJ,  this  edge,  produced,  pierces 
the  meridian  plane  of  rays  OAB  at  0,0'.  Let  OA — O'Q  be  the 
direction  of  the  light,  then  Q,  the  point  where  the  ray  OA — O'Q' 
meets  the  axis  A — A' A",  is  the  intersection  of  this  axis  with  a 
plane  of  rays  through  IJ.  Hence,  any  meridian  plane  of  the 
screw  will  cut  a  straight  element  from  its  surface,  and  a  line  from 
the  plane  of  rays.  This  line  will  pass  through  Q,  and  will  inter- 
sect the  element  in  the  shadow  of  some  point  of  IJ.  Thus,  the 
meridian  plane,  Am,  cuts  from  the  upper  thread  the  line  kj — k'f, 
and  from  the  plane  of  rays,  the  line  raA — m'Q,  which  intersects 
the  former  line  at  I', I,  a  point  of  the  required  shadow. 

In  like  manner,  the  edge  JK  meets  the  meridian  plane  of  rays, 
OAB,  at  a,a'.  Then  through  a,a'  draw  a  ray,  and  Q"  will  be 
determined  as  the  intersection  of  a  plane  of  rays,  through  JK, 
with  the  axis,  A — A' A".  The  meridian  plane,  AB",  now  cuts, 
from  the  edge  JK,  the  point,  B'^B'",  and  from  the  plane  of  rays 
the  line  B"A— B'"Q".  This  line  meets  the  element  ED— 
E'T)"',  cut  from  the  upper  thread,  at  ri",n",  the  shadow  of  some 
point  of  JL. 

Remarks, — a.  The  points,  as  e",e"f,  where  the  shadow  leaves 
the  screw,  at  its  intersection  with  the  helices,  are  all  found  by 
the  indirect  special  method,  called  the  method  of  intersecting  sha- 
dows;  and  which  consists  of  the  following  easy  operations.  Con- 
struct the  shadow  of  the  line,  L,  casting  the  shadow — in  this  case  the 
edge  of  the  nut — upon  any  plane  below.  Find  also,  on  the  same 
plane,  the  shadow  of  M,  the  line  containing  the  required  point  of 
shadow — in  this  case  the  outer  helix.  Then  the  intersection  of 
the  shadow  of  L  with  the  shadow  of  M  will  be  a  point,  through 
which  a  ray  will  meet  M  in  a  point,  m,  and  L,  in  a  point,  I ;  and 
ra  will  be  the  shadow  of  I.  The  point  t,t'  is  thus  found. 

b.  This  construction  is  not  fully  shown  at  R,A",  and  e"e'". 
By  drawing  the  rays  through  these  points,  we  find  the  points 
Y,Y'  and  n,ri  of  which  they  are  the  shadows.     Similarly,  the 
points  which  cast  any  ascertained  point  of  shadow,  may  be  found. 

c.  From  Y,Y'  the  shadow  of  the  nut  falls  on  the  next  thread 
below,  beginning  at  q,q',  the  intersection  of  the  ray  through  Y,Y' 
with  the  shadow  of  the  outer  helix. 

d.  The  point  e,ef  is  the  shadow  of  a,a',  and  is  the  intersection 
of  the  ray  ae — a'e'  with  the  element  be — b"c".     This  one  point  is 
found  by  (41)  as  a  meridian  plane  of  rays  cuts  a  straight  element 
from  a  helicoid. 


62  GENERAL    PROBLEMS. 

2d.  To  find  the  shadow  of  an  assumed  point  of  the  n\it. 

Let  the  point  be  an  angle  of  the  nut,  as  JJ'.  Constructions 
for  the  shadow  of  such  a  point  prove,  on  examination,  to  be 
quite  numerous.  The  following  are  some  of  them. 

First.  By  the  most  simple  direct  method.  Pass  a  vertical  plane 
of  rays  through  J,J'.  Its  intersection  with  the  surface  of  the 
thread,  will  be  a  curve,  easily  formed  by  joining  the  points  in 
which  the  plane  intersects  several  assumed  elements  of  the  screw, 
taken,  for  convenience,  in  the  supposed  vicinity  of  the  desired 
point  of  shadow.  A  ray  through  J,J'  will  meet  this  curve  (not 
shown)  in  the  required  point  of  shadow  W, "W. 

Second.  By  an  obvious  indirect  method.  Find  the  shadows  of 
points,  which  may  be  known  to  be  on  IJ  and  KJ  produced.  The 
indefinite  shadows  of  these  edges,  found  as  in  (66 — Is/),  will  inter- 
sect at  a  point,  W,~W',  which  will  be  the  shadow  of  J, Jx. 

Third.  By  the  following  indirect  method,  interesting  in  theory, 
but  tedious  in  application.  Make  J,J'  the  vertex  of  a  cone, 
generated  by  the  revolution  of  the  ray  through  JJ',  around  a 
vertical  axis  at  JJ'.  The  intersection  of  this  cone  with  the  screw, 
is  easily  found  by  joining  the  points  of  intersection  of  several 
neighboring  elements  of  the  screw  with  the  surface  of  the  cone. 
The  intersection  of  the  ray  element  with  this  curve,  is  its  inter- 
section with  the  screw,  and  therefore  is  the  shadow  of  J, J'. 

Fourth.  Let  a  plane  of  rays  be  passed  through  3,J',  so  as  to 
cut  a  rectilinear  element  from  the  screw.  That  is,  as  such  a 
plane  contains  a  ray  through  JjJ',  it  is  really  required  to  pass  a 
plane  through  a  given  line,  and  so  as  to  cut  a  rectilinear  element 
from  the  screw.  The  given  ray  pierces  the  horizontal  plane, 
W'w',  at  V.  Any  line  through  V  may  be  considered  as  the 
horizontal  trace — on  7t'"w' — of  a  plane  of  rays  through  J,J'. 
The  upper  surface  of  the  upper  thread,  produced,  intersects  the 
plane,  TJ!"W',  in  the  spiral  ivxyz,  and  a  line  from  A  to  any  point 
of  this  spiral  is  the  horizontal  projection  of  an  element  of  this 
thread. 

Next,  make  J,J'  the  vertex  of  a  cone,  generated  by  the  revo- 
lution, about  a  vertical  line  at  J,  J',  of  a  line,  Jt,  which  makes  an 
angle  with  the  plane  Z'"w/,  equal  to  the  angle  made  by  the  ele- 
ments of  the  screw  with  that  plane.  The  circle,  with  radius  Jt, 
and  centre  J,  is  the  base  of  this  cone,  in  the  plane  Z"'«/.  If 
now  a  trace,  VS,  can  be  drawn,  such  that  SJ  and  5A  shall  be 
parallel,  then  these  lines,  being  the  projections  of  lines  of  equal 


SHADES  AND  SHADOWS.  63 

declivity,  will  be  parallel  in  space.  Further,  as  their  tracts,  S, 
and  5,  are  in  the  trace  of  the  plane  of  rays,  the  lines  themselves 
are  in  that  plane,  and  as  5  is  also  a  point  in  the  trace  of  the 
screw,  5A  is  the  element  of  the  screw  which  is  contained  in  a 
plane  of  rays  through  J,J'. 

Now,  to  find  YS,  draw  a  number  of  trial  traces  Y6,  Y7,  etc., 
and  note  the  corresponding  radius  vectors,  6  A,  7 A,  etc.  Then 
draw  J8,  Ji,  etc.,  parallel  to  7 A,  6A,  etc.,  and  note  their  intersec- 
tions, 8,  4,  etc.,  with  the  traces  V7,  V6,  etc.  Then  the  curve, 
8,  4,  etc.,  is  the  locus,  or  place,  of  all  intersections  of  radii  through 
J,  parallel  to  radius  vectors  through  A,  with  the  traces  through 
V.  Hence  S,  where  this  trial  curve,  or  locus,  intersects  the  base, 
/IS,  of  the  cone,  is  the  point  from  which  can  be  drawn  the 
required  trace,  SY,  of  a  plane  of  rays  through  J,  J',  and  cutting  a 
rectilinear  element,  5A,  from  the  screw.  Then  "W, W,  where  the 
ray,  JW — J'W,  meets  this  element,  is  the  required  shadow  of 
J,J'  on  the  screw. 

Remark. — It  is  now  apparent,  from  the  preceding  construction, 
that  it  reduces  itself  to  this  problem  of  two  dimensions.  Through 
a  given  point  (V)  in  the  plane  containing  a  spiral  of  Archimedes 
and  a  circle,  it  is  required  to  draw  a  line  (VS)  intersecting  the 
spiral  and  the  circle  at  points,  from  which  parallels  can  be  drawn, 
respectively,  to  the  pole  of  the  spiral,  and  the  centre  of  the  circle. ^ 

67.  To  find  ike  shadow  of  the  curve  of  shade  on  the  thread  below. 

ls£.   On  any  assumed  element. 

Principles. — This  shadow  is  found  by  the  indirect  special  method 
of  (66 — 1st.  Kem.  a).  Thus,  the  shadow  of  the  curve  of  shade 
upon  the  horizontal  plane,  will  intersect  the  shadow,  on  the  same 
plane,  of  an  assumed  element,  supposed  to  contain  a  point  of 
shadow,  in  a  point,  through  which,  if  a  ray  be  drawn,  it  will 
intersect  both  the  element  and  the  curve  of  shade.  The  former 
intersection  will  be  the  shadow  of  the  latter. 

Construction.— PI.  VII.,  Fig.  22.  To  find  that  point  of  the 
shadow  of  the  curve  of  shade,  hf — /?//,  which  falls  on  the  element 
G"s — GV.  Drawing  rays,  G"X  and  sY,  not  shown  in  the  verti- 
cal projection,  through  G^G',  and  s,s',  the  line  YX  will  be  found, 
as  the  shadow  of  the  element  upon  the  horizontal  plane  of  pro- 
jection. Similarly,  &Z  is  found,  for  the  shadow  of  hf — h'f  on 
the  same  plane.  These  shadows  intersect  at  t.  Then  tu  is  the 
ray  which  intersects  G"s — GV  at  «,w',  which  is  the  required  point 
of  shadow. 


64:  GENERAL  PROBLEMS. 

Remark. — By  producing  the  ray,  that  point  of  hj — h'j',  whose 
shadow  is  tm',  could  be  found. 

2d  On  an  assumed  helix.  This  shadow  is  found  in  the  same 
way  as  was  the  one  just  explained.  Thus:  let  the  outer  helix, 
soe — s'o'e',  be  the  assumed  helix.  VZ  is  a  portion  of  its  indefinite 
shadow,  found  by  connecting  the  shadows  of  several  of  the  points 
of  the  helix,  on  the  horizontal  plane.  This  shadow  intersects 
the  shadow,  7cZ,  of  the  curve  of  shade,  hj — h'j',  at  Z.  Then  the 
ray  Zy — v't'  meets  the  helix  at  v,v'  which  is  the  shadow  of  t' 
(between  h  and  i,  in  horizontal  projection)  on  the  curve  of  shade 
— produced  in  the  present  case. 

3d.  On  another  branch  of  the  curve  of  shade.  Find,  by  (66 — 1st.) 
a  point,  y",  of  the  shadow  of  the  curve  of  shade  hj — h'j'  produced 
upon  an  indefinite  element,  as  Ae,  beyond  b  on  the  supposed 
branch,  a&,  of  the  curve  of  shade.  Join  this  point  with  the  pre- 
viously found  points  of  shadow,  juy",  of  hj — h'j  upon  the  screw. 
The  indefinite  curve  of  shadow  thus  found,  will  intersect  ab  in 
the  required  point,  d,d'. 

Remark. — To  find  how  far  this  curve  of  shadow  of  hj — h'j'  is 
real,  consider  that  it  begins  aty,/,  the  intersection  of  hj — h'j'  with 
the  inner  helix  /GV,  and  is  limited  by  the  ray  through  hthf, 
which  meets  the  shadow,  juy"—j'u'v',  at  p"p'".  Beyond  p"p'",  a 
real  shadow  is  cast,  by  the  outer  helix  through  hh'. 

68.  To  find  the  shadow  of  the  outer  helix  on  the  upper  surface  of 
the  thread  below. 

1st.  On  any  assumed  element.  First  Method. — This  is  the  method 
of  (66— 1st.  Bern,  a)  and  of  (67— 1st).  It  is  shown,  in  PI.  VIII., 
Fig.  23,  in  the  construction  of  the  point  //.  Here  AM— CX/M; 
is  any  assumed  element  produced,  on  which  a  point  of  the  pro- 
posed shadow  may  fall.  MP  is  its  shadow  on  the  horizontal 
plane,  and  B"F  is  the  shadow  of  a  portion  of  that  outer  helix, 
whose  vertical  projection  is  bf"s".  Then  B'/  the  horizontal 
projection  of  a  ray  through  the  intersection  of  these  shadows, 
gives/  which,  projected  in  C"M',  at/',  gives//  as  the  point  of 
shadow  required. 

Second  Method. — This  is  a  special  method  which  may  be  called 
the  method  ly  helical  translation  of  a  plane  of  rays  containing  an 
assumed  element. 

Principles. — When  a  plane  is  perpendicular  to  either  plane  of 
projection,  all  lines  in  it  will  be  projected,  on  that  plane  to  which 
the  given  plane  is  perpendicular,  in  the  trace  of  the  given  plane. 


SHADES  AND  SHADOWS.  65 

Also,  the  intersection  of  the  plane  of  rays  through  the  element, 
with  the  given  helix,  is  the  point  casting  a  shadow  on  that  ele- 
ment, at  its  intersection  with  a  ray  through  that  point. 

Construction.— l\-i  PL  VIII.,  Fig.  23,  let  AT— I"T'  (53—2°)  be 
any  assumed  element.  It  pierces  the  horizontal  plane  at  N',N. 
The  ray  AB — F'B',  through  the  upper  extremity  of  the  element, 
pierces  the  horizontal  plane  at  B',B.  Then  NB  is  the  horizontal 
trace  of  a  plane  of  rays  through  the  element  AN — I"N'.  Let 
this  plane  now  be  revolved  till  it  becomes  perpendicular  to  the 
vertical  plane  of  projection.  Its  line  of  declivity,  AH,  will  then 
be  horizontally  projected  in  AH".  Producing  AH,  the  arc  CU 
corresponds  to  HH".  All  points  of  the  plane  of  rays  revolve 
equally,  therefore  make  T3  =  CU,  and  project  3  at  3'  on  the  outer 
helix.  All  points  of  the  plane  also  rise  equally,  and  at  the  rate 
of  the  rise  of  the  helix,  in  order  that  it  may  still  continue  toxion- 
tain  an  element  of  the  screw.  Therefore  make  I"I"'  equal  to  the 
height  of  3'  above  T',  and  draw  r"3'H"'.  As  the  plane,  NB,  is 
now  perpendicular  to  the  vertical  plane  of  projection,  I'"H'"  is 
its  trace,  and  it  cuts  the  outer  helix  at  u'"u'' '.  In  the  counter 
revolution  of  this  plane,  u'"u"  traverses  a  helical  arc  whose  hori- 
zontal projection  is  equal  to  CU,  hence  make  u"h  =  CU,  project 
h  at  h',  and  then  draw  the  ray  hg — h'g',  which  intersects  the 
assumed  element  AT — A'"!"  at  g,g\  the  required  point  of 
shadow. 

2d.   On  any  helix. 

This  shadow  is  found  by  the  method  of  intersecting  shadows 
(67 — 1st}.  Thus,  let  the  shadow  of  a  spire  of  the  outer  helix  be 
found  on  the  spire  below,  of  the  same  helix.  F  is  the  intersec- 
tion of  the  shadows,  cast  on  the  horizontal  plane,  by  fragments 
of  the  outer  helixes,  near  sf  and  u'.  Then  the  ray  Fs  gives  s, 
whose  vertical  projection  is  s'.  The  same  ray  produced — not 
shown  in  horizontal  projection — gives  u',u.  Therefore  s,s'  on 
the  outer  helix,  is  the  shadow  of  w,w',  a  point  on  the  spire 
above.  In  the  same  way  d",d'"  is  found. 

3d.   On  any  particular  element. 

The  point  cZ,cZ',  in  the  meridian  plane  OA,  is  found  by  the 
ordinary  direct  method,  as  the  intersection  of  a  ray,  Id — b"'df,  with 
the  element  be — b'c'.  2,2'  may  be  found  either  as  //'  was,  or  as 
g,g'  was.  sdd" — s'd'Z'd'"  is  now  the  complete  shadow  of  the 
outer  helix,  through  &,&'",  on  the  lower  thread.  The  visible 
portion  of  an  equal  shadow,  similarly  situated,  is  shown  on  the 

5 


66  GENERAL   PROBLEMS. 

thread  above.     On  the  upper  thread,  the  dotted  edge  of  a 
lar  shadow  is  shown,  as  it  would  appear  if  the  nut  were  re- 
moved. 

69.  The  shadow  of  the  screw  and  nut  on  the  horizontal  plane. 

1st.  The  determination  of  the  lines  casting  this  shadow. 

These  lines  are,  so  much  of  each  of  the  curves  of  shade  as  is 
real,  that  is,  not  concealed  by  a  shadow,  and  on  the  exterior  of 
the  helicoid  ;  so  much  of  each  outer  helix  as  divides  the  illu- 
minated part  of  the  upper  surface  of  a  thread  from  the  unillumiri- 
ated  part  of  the  lower  surface  of  the  same  thread ;  and,  lastly, 
those  edges  of  shade  of  the  nut,  which  do  not  cast  shadows  on 
the  screw. 

2d.  To  construct  the  shadow  of  any  point  in  any  offfie  above  por- 
tions offlie  total  line  of  shade  of  the  screw. 

This  construction  is  made  by  the  simple  method  of  (Prob.  III.). 


PROBLEM  XXIII. 

To  determine,  in  general,  the  shadows  of  a  vertical  right  conoid. 

70.  In  considering  the  shadows  of  the  conoid,  PL  VIII.,  Fig.  25, 
it  is  to  be  remembered,  that,  as  all  the  elements  of  the  conoid 
intersect  the  line  of  striction,  AA" — A',  a  plane  can  be  passed 
through  this  line  and  any  two  parallel  elements,  equidistant  from 
the  meridian  plane,  HK.  All  such  planes  will  have  horizontal 
traces,  parallel  to  AA",  and  intersecting  the  base  of  the  conoid. 
A  tangent  plane  along  the  element  JK,  is  the  extreme  case  of 
the  above  group  of  planes. 

From  the  preceding  it  follows  that  any  plane  of  rays,  passed 
through  the  line  of  striction,  and  whose  trace,  parallel  to  AA", 
is  between  AA"  and  K,  as  at  ER,  will  intersect  the  conoid  in  two 
rectilinear  elements.  Of  these  two  elements,  the  foremost  one, 
as  ET,  will  be  the  indefinite  shadow  of  the  portion,  TA,  of  the 
line  of  striction  upon  the  exterior  of  the  conoid.  This  shadow 
will  be  limited  by  the  ray,  as  A&,  through  A,A'.  The  opposite 
parallel  element,  T"E",  is  the  imnginary  shadow,  on  the  interior 
of  the  conoid,  of  the  portion  TT"  of  the  line  of  striction. 

The  line  Ab  is  also  the  horizontal  projection  of  the  imaginary 
curve  of  shadow,  cast  upon  the  interior  of  the  conoid  by  the  ver- 
tical element  at  A ;  it  being  the  horizontal  projection  of  the 


SHADES  AND  SHADOWS.  67 

intersection  of  a  vertical  plane  of  rays,  through  this  element, 
with  the  conoid. 

"When  the  plane  of  rays,  through  the  line  of  striction,  does  not 
otherwise  intersect  the  conoid,  the  shadow  of  this  line  will  fall 
wholly  on  the  horizontal  plane.  In  the  case  just  considered,  only 
the  portion  T^A"  casts  a  shadow  on  the  horizontal  plane. 

The  remaining  shadows  on  the  horizontal  plane  are  cast  by  the 
curves  of  shade.  They  are  found  by  the  usual  obvious  methods, 

In  this  article,  only  the  lower  nappe  of  the  conoid  has  been 
considered. 


68  GENEKAL   PROBLEMS. 


DIVISION  II. 

SHADES  AND  SHADOWS  ON  DOUBLE  CURVED  SURFACES. 

§  1.— Shades. 

71.  Either  a  line,  or  a  plane,  may  be  placed  tangent  to  a 
double  curved  surface,  at  a  point  of  contact,  only.    Hence  a  point 
in  the  curve  of  shade  of  such  a  surface,  may  be  conceived  of, 
either  as  the  point  of  contact  of  a  tangent  ray,  or  of  a  tangent  plane 
of  rays. 

72.  A  tangent  plane  of  rays  is,  practically,  conceived  of  as  con- 
taining some  given  ray,  which,  with  the  point  of  contact,  deter- 
mines that  tangent  plane.     A  tangent  ray  is  practically  conceived 
of  as  a  tangent  to  some  given,  or  constructed,  section  of  the  given 
double-curved  surface. 

73.  From  the  two  preceding  articles,  we  have  the  two  follow- 
ing general  methods  for  finding  points  in  the  curve  of  shade  of  any 
double-curved  surface. 

First  general  method. — That  of  tangent  planes  of  rays.  By  any 
of  the  solutions  of  the  problem  of  orthographic  projections, 
requiring  a  plane  to  be  drawn  through  a  given  line  and  tangent 
to  a  double-curved  surface,  construct  a  tangent  plane  through  any 
assumed  ray.  Its  point  of  contact  with  the.  given  double-curved 
surface,  will  be  a  point  in  the  curve  of  shade  of  that  surface. 

74.  Second  general  method. — That  of  tangent  rays.     Construct 
any  simple  section  of  the  given  surface,  whose  plane  shall  contain  a 
ray  of  light.     Draw  a  parallel  ray,  tangent  to  this  section,  and  its 
point  of  contact  ivill  be  a  point  of  the  required  curve  of  shade. 

Remarlcs. — a.  The  first  of  these  methods  has  no  formal  illustra- 
tions among  the  following  problems,  numerous  special  methods 
being  more  convenient  in  the  various  cases. 

b.  The  second  general  method  is  convenient,  in  finding  certain 
extreme  points,  as  the  highest  and  lowest,  on  double-curved  sur- 
faces of  revolution.  It,  therefore,  has  no  separate  formal  illustra- 
tions in  the  succeeding  problems. 


SHADES  AND  SHADOWS.  69 

c.  Each  of  the  following  problems  is  solved  by  a  special  method, 
which  is  convenient  for  the  purpose,  but  is  not  necessarily  appli- 
cable only  to  the  problem  thus  solved. 

75.  First  special  method. — That  of  normal  planes  of  shade.  The 
essential  principle  of  this  method,  which  is  applicable  only  to  the 
sphere,  is,  that  the  plane  of  the  curve  of  shade — briefly  called  the 
plane  of  shade — of  the  sphere,  is  perpendicular  to  the  direction 
of  the  light. 

The  peculiarity  and  convenience  of  this  method  is,  that  it 
requires  but  one  projection  of  the  sphere,  as  •will  now  be 
shown. 


PROBLEM  XXIY. 

To  find  the  curve  of  shade  of  the  sphere,  using  only  one  projection 

of  the  sphere. 

Principles. — Let  the  vertical  projection,  only,  be  given,  and 
let  the  vertical  plane  of  projection  be  taken  through  the  centre 
of  the  sphere. 

Let  that  plane  of  rays,  which  is  perpendicular  to  the  vertical 
plane  of  projection,  be  called  the  perpendicular  plane  of  rays, 
and  let  the  one,  whose  angle  with  the  vertical  plane  equals  that 
made  by  the  light  with  that  plane,  be  called  the  oblique  plane 
of  rays.  Let  this  oblique  plane  of  rays  contain  the  centre  of 
the  sphere. 

The  oblique  plane  of  rays,  and  the  plane  of  shade,  are  per- 
pendicular to  each  other,  and  to  the  perpendicular  plane  of  rays, 
hence  their  traces  on  the  latter  plane  will  be  perpendicular  to  each 
other  at  the  centre  of  the  sphere,  and  their  common  vertical 
trace,  T,  will  be  perpendicular  to  the  trace  of  the  perpendicular 
plane  of  rays,  and  will  pass  through  the  centre  of  the  sphere. 

Now  let  the  perpendicular  plane  be  revolved  about  its  trace, 
and  into  the  vertical  plane  of  projection.  Then  consider  the 
great  circle,  which  is  the  vertical  projection  of  the  sphere,  as  the 
circle  of  shade,  revolved  about  its  diameter,  T,  into  the  vertical 
plane.  Project  any  of  its  points  into  the  trace  of  the  perpen- 
dicular plane.  Eevolve  it  into  the  trace  of  the  plane  of  shade 
on  the  perpendicular  plane.  Counter  revolve  the  latter  plane 
to  its  primitive  position,  carrying  the  revolved  point.  Counter 


70  GENERAL  PROBLEMS 

project  the  point  to  the  parallel  plane  from  which  it  originally 
came. 

Construction. — PI.  IX.,  Fig.  26.  LPKE  is  the  vertical  pro 
jection  of  the  sphere.  LK  is  the  vertical  projection  of  a  ray, 
and  the  vertical  trace  of  the  perpendicular  plane  of  rays  through 
the  centre  of  the  sphere.  Or,  assumed  at  pleasure,  is  the 
revolved  position  of  a  ray,  about  LK  as  an  axis,  and  is  also  the 
revolved  position  of  the  trace  of  the  oblique  plane  of  rays,  on 
the  perpendicular  plane  of  rays,  LK.  OS,  perpendicular  to  rO, 
is  the  revolved  position,  about  the  trace  LK  as  an  axis,  of  the 
trace  of  the  plane  of  shade  on  the  plane  LK.  POE  is  the  com- 
mon vertical  trace  of  the  oblique  plane  of  rays,  rO,  and  the 
plane  of  shade,  OS,  on  the  vertical  plane  of  projection.  It  is 
perpendicular  to  LK,  and  is  the  transverse  axis  of  the  ellipse 
into  which  the  circle  of  shade  is  projected.  LRS  is  the  circle 
of  shade,  after  revolution  about  POR,  and  into  the  vertical 
plane  of  projection.  Assume  any  point  5,  project  it,  at  &',  into 
the  trace  LK.  Revolve  it,  as  at  b'b",  about  O  as  a  centre,  into 
the  trace  OS  of  the  plane  of  shade.  Counter  revolve  Z>",  with 
the  plane  LK  to  V"  in  the  primitive  position  of  the  plane  LK, 
and  counter  project  V"  to  B,  in  the  plane  db  whence  it  came. 
Then  B  will  be  a  point  in  the  required  projection  of  the  curve 
of  shade  of  the  given  sphere. 

All  the  points  of  this  curve  may  be  found  as  just  described, 
but  it  is  not  necessary  to  find  more  than  a  quadrant,  REs,  of  the 
curve,  in  this  way.  For,  taking  parallel  planes,  as  db  and  CA, 
equidistant  from  LK,  the  point  V"  can  also  be  counter  projected 
to  C,  a  point  of  shade  opposite  to  B  from  the  axis  sk.  Then  r. 
and  B ;  and  h  and  C,  etc.,  will  be  equidistant  from  the  transverse 
axis  PR. 

76.  For  the  practical  case  in  which  the  projections  of  a  ray 
make  angles  of  45°  with  the  ground  line,  the  angle,  as  rOL. 
made  by  the  ray  with  the  vertical  plane  of  projection,  will  be 
35°-16'  (21).  OS,  being  perpendicular  to  Or,  KOS  will  be  an 
angle  of  54°-44:'.  This  angle  may  easily  be  constructed,  then  it 
will  not  be  necessary  to  draw  Or.  Thus,  take  on  OK  any  dis 
tance,  Os,  and  find  the  hypothenuse  of  a  right  angled  triangle, 
each  of  whose  other  sides  equals  Os.  On  a  perpendicular,  a*» 
«S,  at  s,  lay  off  this  hypothenuse,  then  SO,  the  new  hypothenuse 
thus  found,  will  make  the  required  angle  of  54°-44'  with  OK, 
and  will  therefore  be  the  true  position,  after  revolution  of  the 


SHADES  AND  SHADOWS.  71 

plane  LK,  of  the  trace  of  the  plane  of  shade  upon  the  plane 
LK. 

77.  Second  special  method.  That  of  the  intersection  of  the  plane 
of  shade  with  the  given  double  curved  surface.  This  method  can 
be  applied  to  all  the  double  curved  surfaces  of  the  second 
order  (Des.  Geom.)  for  the  curve  of  contact  of  a  cylinder,  or 
cone,  of  rays,  with  such  a  surface,  is  a  plane  curve  of  shade.  This 
method  consists  principally  in  making  an  auxiliary  rectilinear  projec- 
tion of 'the  curve  of  shade,  which  may  be  done  by  talcing  an  auxiliary 
vertical  plane  of  projection  perpendicular  to  that  plane  of  shade. 


PROBLEM  XXY. 
To  find  the  curve  of  shade  on  an  ellipsoid  of  revolution. 

Principles. — Let  the  ellipsoid  be  a  prolate  spheroid  (Des. 
Geom.  219)  and  let  its  principal  axis  be  vertical.  Any  auxiliary 
horizontal  plane  will  then  cut  a  horizontal  line  from  the  plane 
of  shade  and  a  circle  from  the  ellipsoid. 

The  two  points  of  intersection  of  these  two  lines  will  be 
two  points  in  the  curve  of  shade. 

Construction.—  PL  IX.,  Fig.  27.  Only  half  of  the  ellipsoid  is 
here  shown.  A — AT)'  is  its  semi-principal  axis.  BC — B'C'  is 
the  diameter  of  its  greatest  horizontal  section.  E^F",  parallel 
to  LA,  the  horizontal  projection  of  a  ray  of  light,  is  the  ground 
line  of  a  new  vertical  plane  of  projection,  parallel  to  the  light. 
The  projection  of  the  semi-ellipsoid  on  this  plane  is  a  semi-ellipse 
equal  to  B'D'C'.  Having  noted  L  and  L',  two  projections  of 
any  point  in  the  ray  LA — I/ A',  project  A  at  A",  and  L  at  L", 
making  ~L"h"  equal  to  L7i'.  Then  L"A"  is  the  projection  of 
the  light  on  the  new  plane  of  projection.  Next,  draw  the  tan- 
gent ray  at  T",  and  T"A"  will  be  the  auxiliary  rectilinear 
projection  of  the  curve  of  shade.  Now  the  auxiliary  horizontal 
plane,  m"n",  taken  at  pleasure,  cuts  from  the  plane  of  shade  the 
line  a" — ab,  and  from  the  ellipsoid,  the  circle  m"n" — anb. 
These  intersect  at  a,a"  and  &,cr",  two  points  of  the  curve  of 
shade.  On  the  primitive  vertical  projection,  these  points  are 
projected  at  a'  and  &',  on  a  line  a'b',  as  far  from  B'C'  as  a"  is 
from  the  ground  line  E"F." 


72  GENERAL  PROBLEMS. 

Similarly  other  points  are  found.  A."  is  projected  at  S  and  V, 
and  then  at  S'  and  V.  T"  is  likewise  projected  at  T  and  T's 
Drawing  a  tangent,  parallel  to  I/ A',  we  find  t',t.  on  the  meridian 
curve  parallel  to  the  vertical  plane  whose  ground  line  is  B'C'. 

Only  the  portion  t'b'VC'  of  the  shade  is  visible  in  vertical 
projection.  The  boundary  of  the  invisible  portion  of  the  shade 
is  dotted. 

Remark. — The  point  T"  is  exactly  constructed  on  the  princi- 
ple that  the  diameter  of  an  ellipse,  through  the  point  of  tangency, 
bisects  all  the  chords  of  the  ellipse  parallel  to  the  tangent. 
Hence  take  any  two  chords,  parallel  to  L"A",  bisect  them,  and 
draw  a  line  through  their  middle  points.  Such  a  line  will  meet 
the  circumference  at  T",  the  point  of  tangency  of  a  tangent 
parallel  to  L"A". 

78.  Third  special  method.  This  is  essentially  the  second  general 
method  (74)  but  modified  so  as  to  be  most  conveniently  applica- 
ble to  the  construction  of  the  curve  of  shade  on  an  ellipsoid  of 
three  unequal  axes. 

A  plane  of  rays  passed  through  the  mean  axis,  will  be  per- 
pendicular to  the  plane  of  the  longest  and  shortest  axes.  Make 
the  ellipse,  E,  cut  from  the  ellipsoid  by  this  plane,  the  base  of  a  cylin- 
der whose  right  section  shall  be  a  circle.  Such  a  cylinder  may  be 
considered,  as  the  projecting  cylinder  of  the  ellipse  upon  the  plane  of 
right  section  of  the  cylinder,  taken  as  a  new  plane  of  projection. 
Other  parallel  planes  of  rays  will  cut  ellipses,  similar  to  ~E,from  the 
ellipsoid.  Their  projections  on  the  new  plane  of  projection  will 
therefore  be  circles,  to  which  tangent  rays  may  be  drawn,  which  will 
give  projections  of  points  of  the  curve  of  shade. 


PROBLEM  XXVI. 

To  find  die  projections  of  the  curve  of  shade  on  an  ellipsoid  of  fliree 

unequal  axes. 

Principles. — Let  the  longest  axis  be  vertical,  and  the  plane  of 
the  longest  and  shortest  axes,  parallel  to  the  vertical  plane  of 
projection.  To  locate  the  projecting  cylinder  described  in  (78) 
make  the  centre  of  the  ellipsoid  the  centre  of  an  auxiliary 
sphere,  whose  radius  shall  be  equal  to  the  semi-mean  axis  of  the 


SHADES  AND  SHADOWS.  73 

ellipsoid  ;  then  the  cylinder  will  be  tangent  to  this  sphere  .in  a 
circle,  which  may  ba  considered  as  a  circular  projection  of  the 
elliptical  base  of  the  cylinder. 

Any  plane,  parallel  to  the  circle  just  described,  may  be  taken 
as  a  new  horizontal  plane  of  projection,  on  which  a  new  hori- 
zontal projection  of  a  ray  of  light  must  be  constructed. 

Construction.—  PI.  IX.,  Fig.  28.  0,O'  is  the  centre  of  the 
ellipsoid.  Let  O'D'  be  the  length  of  the  semi-mean  axis,  whose 
vertical  projection  is  0'.  A  circle,  with  radius  O'D'  and  centre 
0',  is  the  vertical  projection,  partially  shown,  of  the  auxiliary 
sphere  above  described. 

LO  and  I/O'  are  the  projections  of  a  ray  of  light.  Then 
L'F'  is  the  vertical  trace  of  a  plane  of  rays,  perpendicular,  here, 
to  the  vertical  plane  of  projection,  in  being  perpendicular  to  the 
plane  of  the  longest  and  shortest  axes.  J'F'  is  the  vertical  pro- 
jection of  the  ellipse  contained  in  this  plane  of  rays,  and  tan- 
gents, from  J'  and  F',  to  the  vertical  projection  of  the  auxiliary 
sphere,  as  at  e',  are  the  extreme  elements  of  its  projecting  cylinder. 
Take,  now,  any  new  ground  line,  MN,  parallel  to  OV,  and  make 
O'O"  perpendicular  to  it,  and  make  0"  and  O  equidistant  from 
their  respective  ground  lines,  MN  and  MP.  Likewise  make  L" 
in  a  perpendicular  from  L'  to  MN,  and  as  far  from  MN  as  L  is 
from  MP.  Then  L^O"  is  the  new  horizontal  projection  of  a  ray 
of  light.  Now  the  circle  with  centre  O"  and  radius  0"F  (=OY} 
is  the  horizontal  projection  of  the  section  J'F',  and  its  diameter 
/h,  perpendicular  to  L"O",  determines  precisely  the  points  of 
contact,  f  and  n,  of  rays  tangent  to  the  section  J'F' — JnYf. 
These  points  being  projected  at/'  and  n',  give  f,f  and  n,nf  for 
two  points  of  the  required  curve  of  shade. 

By  drawing  the  tangent  rays  at  t'  and/I",  we  find  the  line  t'T', 
which  bisects  all  the  chords,  as  K'B',  parallel  to  the  tangents  at 
t'  and  T'.  These  chords  are  the  vertical  projections  of  ellipses, 
whose  centres,  as  ra',  are  in  the  line  2'T',  and  whose  horizontal 
projections  are  circles  having  their  centres  in  T£,  the  horizontal 
projection  of  t'T'. 

Thus,  m  is  the  centre,  and  mK  the  radius,  of  the  horizontal 
projection  of  K'B',  and  kb,  its  diameter  perpendicular  to  L"O", 
determines  the  points  of  shade,  k  and  5,  vertically  projected  at 
k'  and  V.  In  the  same  way  g,g'  and  hji'  are  found. 

Through  the  points  now  found,  the  projections  of  the  required 
curve  can  be  sketched  ;  showing  that  it  will  be  included  between 


74:  GENERAL  PROBLEMS. 

tangents  (not  shown)  as  at  u  and  y,  tangent  also  to  the  vertical 
projection,  t'f'T'u,  and  parallel  to  O'O". 

The  horizontal  projection  of  the  curve  of  shade  is  also  tangent, 
at  u  and  y,  to  the  ellipse,  notshown,  whose  minor  axis  is  Oa — 0', 
and  whose  longer  axis  is  the  distance,  on  KF,  between  two 
lines,  perpendicular  to  it,  and  tangent  to  the  ellipse  T'C'A'J' ; 
and  the  former  ellipse  is  the  new  horizontal  projection  of  the 
ellipsoid. 

The  visible  portion  of  the  shade  is  shaded  in  vertical  pro- 
jection. 

79.  fourth  special  method.     This  is  essentially  the    special 
method  of  (51),  but  modified  so  as  to  be  also  applicable  to  the 
ellipsoid  of  three  unequal  axes.     It  is  a  property  of  these  ellip- 
soids— employed  in  the  preceding  article — that  a  set  of  parallel 
sections  of  any  one  of  them,  is  composed  of  similar  ellipses. 
Add  to  this,   as  just  previously  shown,  that  a  diameter  can 
obviously  be  found,  in  the  plane  of  the  longest  and  shortest 
axes,  which  shall  be  equal  to  the  mean  axis.     The  plane  of  this 
diameter  and  mean  axis  will  evidently  intersect  the  ellipsoid  in  a 
circle,  and  so  will  all  parallel  planes.     Now  by  taking  a  plane 
of  projection — treated  most  conveniently  as  a  new  horizontal 
plane — and  made  parallel  to  these  circles,  we  can  readily  con- 
struct the  elements  of  shade  on  a  series  of  auxiliary  tangent 
cones  having  these  circles  for  their  bases.     The  intersection  of 
the  elements  of  shade  with  the  circles  will  be  points  in  the  curve 
of  shade  of  the  ellipsoid. 

The  construction  is  left  for  the  student  to  make. 

80.  fifth  special  method.     This  is  the  method  of  projections  of 
rays  on  meridian  planes.     Its  essential  principle  is,  that  the  point 
of  contact  of  a  tangent  line  to  any  meridian  curve  of  a  surface 
of  revolution,  and  parallel  to  the  projection  of  a  ray  on  the 
plane  of  that  curve,  is  a  point  of  the  curve  of  shade  of  that  sur- 
face.    For,  a  point  of  shade  is  the  point  of  contact  of  a  tangent 
plane  of  rays  (73).     But  such  a  plane  is  perpendicular  to  the 
meridian  plane  through  its  point  of  contact.     Hence  its  trace  on 
the  meridian  plane  is  the  projection  of  a  ray  upon  that  plane,  and 
is  also  a  tangent  to  the  meridian  curve,  at  the  point  of  contact 
of  the  tangent  plane  ;  that  is,  at  a  point  of  shade  ;  which  proves 
the  principle  above  stated. 


SHADES  AND   SHADOWS.  75 

The  general  operations  of  solution  under  this  method   are 
exemplified  as  follows : 


PROBLEM  XXVII. 
To  construct  the  curve  of  shade  upon  a  torus. 

Principles. — Conceive  of  a  rectangle,  with  one  of  its  shorter 
sides  made  the  diameter  of  a  semi-circle.  If  the  entire  plane 
figure,  thus  formed,  be  revolved  about  the  opposite  side  of  the 
rectangle,  the  solid  generated  will  be  a  torus. 

If  the  axis  of  the  torus  be  supposed  to  be  vertical,  the  points 
of  shade  on  its  greatest  horizontal  section  will  be  the  points  of 
contact  of  two  parallel  and  vertical  tangent  planes  of  rays.  The 
points  of  shade  on  the  meridian  curve  parallel  to  the  vertical 
plane,  will  be  the  points  of  contact  of  two  planes  of  rajs  per- 
pendicular to  the  vertical  plane  of  projection. 

Hence,  in  finding  the  four  points  just  described,  we  make  a 
rudimentary  application  of  the  first  general  method  (73).  Other 
points  of  shade  are  mostly  found  by  the  fifth  special  method  (80). 

Constructions. — 1°.  To  find  the  four  points  on  the  visible  bounda- 
ries of  the  torus.  PI.  IX.,  Fig.  29.  Let  ATB*  and  A'L'B'  be 
the  projections  of  the  torus,  and  RL — R'L'  a  ray  of  light.  Then, 
by  drawing  T/,  perpendicular  to  RL,  T  and  t'  will  be  determined 
as  the  horizontal  projections  of  the  points  of  contact  of  vertical 
planes  of  rays,  tangent  to  the  greatest  circle  of  the  torus  at  T,T' 
and  t.t '. 

Likewise,  at  C'  and  D',  the  centres  of  the  semi-circular  ends 
of  the  elevation,  draw  C'n'  and  D'c'.  perpendicular  to  R'L',  and 
project  n'  at  n,  and  c'  at  c,  then  n,n'  and  c,<?'  will  be  the  points 
of  shade  on  that  meridian  section  of  the  torus,  which  is  parallel 
to  the  vertical  plane  of  projection. 

2°.  The  highest  and  lowest  points.  These  are  in  the  meridian 
plane  of  rays  KL.  Revolve  this  plane,  with  the  ray  RL — R'L', 
about  a  vertical  axis  at  L,  till  parallel  to  the  vertical  plane  of 
projection.  Then  r"L — R"L'  is  the  revolved  position  of  the 
ray  RL— R'L'.  Through  C'  and  D',  draw  CV  and  DV  per- 
pendicular to  R"L",  and  project  u'  at  u  and  a!  at  a.  Then  a,a' 
and  u,uf  are  the  revolved  positions  of  the  highest  and  lowest 
points.  In  the  counter  revolution,  a,a'  returns  to  a'V,  and  uju! 


76  GENERAL   PROBLEMS. 

to  u'V",  vhich  are  the  highest  and  lowest  points  of  the  required 
carve  of  shade.  These  points  are  thus  found  by  a  simple  appli- 
cation of  the  second  general  method  (74). 

3°.  To  find  intermediate  points.  Assume  any  meridian  plane, 
as  d"Le",  project  the  ray  RL — R'L'  upon  it  by  projecting  lines, 
as  Rr — RV,  perpendicular  to  d"~Le,".  L,I/,  being  in  the  assumed 
plane,  is  its  own  projection  on  that  plane,  and  rL  and  r'~L'  are 
the  projections  of  the  projection  of  the  ray,  RL — R'L',  upon  the 
plane  d"Le". 

If,  now,  the  plane  d""Le"  be  revolved  about  a  vertical  axis  at 
L,  till  it  is  parallel  to  the  vertical  plane  of  projection,  the  meri- 
dian section  contained  in  it  will  coincide  with  the  vertical  pro- 
jection of  the  torus,  and  the  projected  ray  rL — r'\J  will  appear 
at  R""L— R'"L'.  From  C'  and  D',  draw  lines,  Q'd'  and  DV, 
perpendicular  to  the  revolved  ray  R///7L — R//XI/,  and  project  d' 
and  e'  at  d  and  e.  Then,  in  the  counter  revolution  of  the  meri- 
dian plane,  d,d'  will  return  in  a  horizontal  arc  to  d",d'",  and  e,e' 
to  e",e'",  which  are  points  of  the  required  curve  of  shade. 

Other  points  can  be  similarly  found. 

81.  The  special  method  of  (51)  may  be  applied  to  any  double 
curved  surface  of  revolution,  but  for  the  sake  of  completeness 
of  illustration,  and  to  show  how  it  would  be  applied  in  finding 
the  curve  of  shade  on  a  warped  hyperboloid,  it  is  here  applied 
to  a  concave  double  curved  surface  of  revolution,  analogous  in 
form  to  that  warped  surface.  This  method  will  therefore  be  next 
described  in  immediate  connexion  with  the  "  construction  "  of  the 
following  example : 


PROBLEM  XXVIII. 
To  find  the  curve  of  shade  on  a  piedouche. 

Principles. — A  Piedouche,  PI.  X.,  Figs.  30,  31,  is  a  little  orna- 
mental pedestal,  used  for  the  support  of  a  bust,  a  statuette,  etc. 

Conceive  of  an  ellipse,  whose  axes  are  in  a  vertical  plane,  but 
oblique  to  the  horizontal  plane.  Suppose  a  vertical  line,  L,  in 
the  plane  of  the  ellipse,  and  so  placed  as  to  converge  downwards 
towards  its  minor  axis.  Then  let  that  arc  of  the  ellipse  which  is 
included  between  horizontal  tangents,  and  convex  towards  L,  be 
revolved  about  L  as  an  axis,  and  a  concave  double  curved  sur- 


&HADES  AND  SHADOWS.  77 

face  of  revolution  will  be  formed,  which  will  be  the  principal 
surface  of  the  piedouche.  The  lower  base  will  evidently  be 
larger  than  the  upper,  and  both  may  be  made  the  bases  of  short- 
cylinders  having  L  for  their  axis.  See  PI.  X.,  Fig.  30,  which 
shows  how  the  generating  semi-ellipse,  AEB,  may  be  formed 
from  a  semi-circle,  as  ACB,  where  DE,  etc.  —  DC,  etc. 

The  elliptical  generatrix  may  be  replaced  by  a  compound 
curve,  composed  of  circular  arcs,  tangent  to  each  other,  and  with 
different  radii. 

Remark. — The  piedouche  formed  by  the  second  method  will 
answer  every  graphical  purpose,  but  is  less  ornamental,  since  a 
circle  is  a  monotonous  curve,  its  points  being  equidistant  from 
the  centre;  while  the  gradually  varying  distance  of  the  points 
of  an  ellipse  from  its  centre  gratifies  the  natural  love  of  variety. 
While  dwelling  on  a  point  of  taste,  it  may  be  added  that  the 
generatrix  should  begin  and  end  on  horizontal  tangents,  because 
it  thus  suggests  completeness  •  and  this,  because  horizontal  boun- 
daries imply  stability  ;  and  this,  again,  because  gravity,  in  refer- 
ence to  whose  action  the  piedouche  is  to  be  stable,  acts  vertically. 

Constructions. — 1°.  Of  the  piedouche,  and  of  some  preliminary 
points  of  shade.  PI.  X.,  Fig.  31.  Assume  the  bases,  H'C'  and 
F'G",  of  the  double  curved  portion  of  the  piedouche.  At  F', 
take  any  vertical  height  F'E',  greater  than  half  the  distance 
between  the  assumed  bases.  Through  E',  draw  the  line  E'l', 
parallel  to  the  bases,  and,  from  C',  let  fall  C'D',  perpendicular  to 
E'l'.  Then,  with  E'F'  as  a  radius,  describe  the  quadrant  from 
F'  to  E'l',  and  with  D'C'  as  a  radius,  describe  the  quadrant  from 
ET  to  C'. 

The  compound  curve,  thus  formed,  will  be  the  meridian  curve, 
whose  revolution  about  a  vertical  axis,  A — T'A',  will  generate 
the  double  curved  portion  of  the  piedouche.  Its  cylindrical 
bases  are,  as  shown  in  vertical  projection,  above  and  below  the 
double  curved  portion.  These  bases  are  horizontally  projected 
in  the  concentric  circles,  A — IIYB,  and  A — GWF. 

In  applying  the  method  of  (51)  some  of  the  assumed  circles 
of  contact  of  the  auxiliary  surface  may  prove  to  be  beyond  the 
limits  of  the  curve  of  shade,  so  that  it  is  best  to  construct  the 
highest  and  lowest  points  of  that  curve  first,  as  in  (Prob. 
XXVII.). 

Let  AJ — A'J'  be  a  projection  of  a  ray  of  light.  Eevolve 
this  ray  about  the  axis  A — T'A'  to  the  position  AJ" — A'J'" 


78  GENERAL    PROBLEMS. 

\ 

Draw  IV  and  EV'"  perpendicular  to  A'J"',  then  u'  and  r'", 
horizontally  projected  at  u  and  r",  will  be  revolved  positions  of 
the  highest  and  lowest  points.  In  the  counter-revolution,  these 
points  return  to  IT,!!'  and  r,r',  which  are  their  real  positions. 

By  drawing  IV  and  EV,  perpendicular  to  A'J',  we  find  a',a 
and  n',71,  the  points  of  tangency  of  planes  of  rays  perpendicular 
to  the  vertical  plane  of  projection.  That  is,  aa!  and  nn'  are  the 
points  of  shade  on  the  meridian  curves  forming  the  vertical  pro- 
jection of  the  piedouche,  and  contained  in  the  meridian  plane 
HAF,  parallel  to  the  vertical  plane  of  projection.  See  the  first 
general  method  (73). 

2°.  Of  intermediate  points  of  shade,  first  illustration  of  (81) 
Principles.  Assume  any  horizontal  section  of  the  piedouche — 
such  a  section  being  circular  because  the  axis  is  vertical — and 
make  it  the  circle  of  contact,  C,  of  a  tangent  cone,  whose  ele- 
ments will  therefore  be  tangent  to  the  meridian  curves  of  the 
piedouche.  Drawing  planes  of  rays,  tangent  to  this  cone,  we 
find  its  elements  of  shade.  The  two  points  of  intersection  of 
these  elements  with  the  circle  of  contact,  C,  will  be  points  of 
shade  on  the  piedouche  (51).  At  the  circle  of  the  gorge,  the 
cone  will  become  a  tangent  cylinder. 

Construction. — Assume  any  horizontal  plane,  K'L',  PI.  X,  Fig. 
31,  between  the  highest  and  lowest  points.  At  either  point,  as 
I/,  of  its  intersection  with  the  meridian  boundary  of  the  vertical 
projection  of  the  piedouche,  draw  the  tangent  L'S'.  This  tangent 
will  be  an  element  of  the  auxiliary  tangent  cone,  whose  base  is 
Z'L'. 

The  circle,  with  radius  A7c,  equal  to  X'L',  is  the  horizontal 
projection  of  the  circle  of  contact,  or  base,  of  the  cone,  and  A,S' 
is  the  vertex  of  this  cone.  The  ray  AK — S'K',  through  the 
vertex  A,S',  pierces  the  plane,  KT/,  of  the  base  at  K',K.  Hence 
K&  is  the  horizontal  projection  of  the  trace  on  the  plane  K'L', 
of  a  plane  of  rays,  tangent  to  the  auxiliary  cone.  Project  k  at  k' ; 
then,  as  k,k'  is  the  intersection  of  the  element  of  shade,  Ak — 
S'&',  of  the  cone,  with  the  circle  of  contact,  Z'L',  it  is  a  point  of 
tangency  of  the  plane  of  rays,  K&,  with  the  piedouche ;  that  is, 
k,k'  is  a  point  in  the  curve  of  shade  of  the  piedouche.  But  as 
two  planes  of  rays  can  be  drawn,  tangent  to  the  cone,  each  cone 
will  give  two  points  of  shade.  Thus,  by  drawing  the  chord  kt, 
perpendicular  to  KA,  we  find  the  point  of  shade,  t}t'}  both  of 
whose  projections  are  invisible. 


SHADES  AND  SHADOWS.  79 

Likewise,  by  assuming  a  circle  of  contact  5V,  whose  radius,  in 
horizontal  projection,  is  AQ,  equal  to  half  of  Z/c',  and  which  is 
the  base  of  a  tangent  cone  whose  vertex  is  W,  we  find  Q,Q'  and 
R,R'  for  two  more  points  of  the  curve  of  shade. 

Finally,  at  the  circle  of  the  gorge, /'m',  horizontally  projected 
with  a  radius  A/z,  equal  to  half  of  f'm',  the  auxiliary  cone  becomes 
a  cylinder.  Hence,  by  drawing  the  diameter  gh,  perpendicular 
to  AK,  the  direction  of  the  light,  we  find  the  two  points  of  shade, 
g,gf  and  h,h'. 

Remark. — The  elements  of  shade,  kA  and  £A,  produced  in 
the  direction  kA  and  tA,  will  meet  the  base  of  a  cone,  tangent 
to  the  lower  nappe  of  the  piedouche,  and  having  the  same  inclina- 
tion that  the  element  I/S'  has,  and  will  therefore  give  two  more 
points  of  shade,  which  can  be  easily  found. 

3°.  Second  illustration  of  the  method  of  (51). 

Principles. — Make  any  circle,  between  the  highest  and  lowest 
points  of  shade,  the  circle  of  contact  of  an  auxiliary  sphere,  tan- 
gent to  the  piedouche  internally.  The  meridian  plane,  parallel 
to  the  vertical  plane  of  projection,  and  the  plane  of  the  assumed 
circle  of  contact,  which  is  horizontal,  may  be  taken  in  this  con- 
struction as  the  planes  of  projection.  But  the  plane  of  the  curve 
of  shade  of  the  sphere  is  perpendicular  to  the  direction  of  the 
light,  hence  its  traces  will  be  perpendicular  to  the  projections  of 
a  ray.  The  -vertical  trace  will  pass  through  the  centre  of  the 
sphere,  and  its  intersection  with  that  diameter  of  the  circle  of 
contact  which  lies  in  the  meridian  plane  just  mentioned,  is  a  point 
of  the  horizontal  trace  of  the  plane  of  shade  of  the  sphere.  The 
intersections  of  this  horizontal  trace  with  the  circumference  of 
the  assumed  circle  of  contact,  will  be  two  points  of  the  curve  of 
shade  of  the  piedouche,  being  the  points  in  which  the  curve 
of  shade  of  the  sphere  intersects  the  circle  of  contact  of  the  sphere 
and  piedouche  (51). 

Construction. — PI.  X,  Fig.  31.  Let  M'N' — dNe  be  the  assumed 
circle  of  contact.  Produce  E'N'  to  T'  in  the  axis  of  the  piedouche ; 
then  T'N'  is  the  radius  of  the  auxiliary  sphere,  whose*circle  of 
contact  with  the  piedouche  is  M'N'.  Then  T'O',  perpendicular 
to  AM',  is  the  vertical  trace  of  the  plane  of  shade  of  the  sphere, 
and  O',O  is  a  point  of  its  horizontal  trace,  eOd.  The  points  e,e' 
and  d,d',  as  explained  above,  are  therefore  points  of  the  curve  of 
shade  of  the  piedouche.  Through  the  points  now  found,  the 
curve  can  be  sketched.  It  is  wholly  invisible  in  horizontal  pro- 


80  GENERAL   PROBLEMS. 

jection.     That  part  is  visible  in  vertical  projection  which  is  in 
front  of  the  meridian  plane  GAF. 

Remarks. — a.  The  curve  of  shade  on  a  sphere  may  be  found  as 
in  Prob.  XXIV.,  in  Prob.  XXV.,  in  Prob.  XXVfl.  Its  points 
may  also  all  be  found  as  are  the  highest  and  lowest  on  the 
piedouche  or  torus,  since  a  set  of  vertical  planes  of  rays  would 
cut  the  sphere  in  circles ;  or  they  may  be  found  by  auxiliary 
tangent  cones,  as  in  (2°.) 

b.  Since  the  piedouche  is  a  surface  of  revolution,  its  curve  of 
shade  may  be  found  as  in  the  problem  of  the  torus.    Likewise  the 
curve  of  shade  of  the  torus  may  be  found  by  the  methods  just 
employed.     In  fact,  these  methods  are  especially  applicable  to  the 
interior  half  of  the  annular  torus,  a  surface  generated  by  the 
revolution  of  a  circle  about  a  line  exterior  to  it,  but  in  its  plane. 

c.  By  inspection  of  the  figure  of  the  piedouche,  the  following 
curious  property  may  be  observed.     Near  the  points  R,R',  Q,Q', 
1,1',  and  2,2',  tangent  rays  may  be  drawn  to  the  projections  of 
the  curve  of  shade  at  the  two  projections  of  the  same   point. 
Hence  such  tangent  rays  are  really  tangent  to  the  curve  of  shade 
in  space.     These  tangents  evidently  include  the  greatest  hori- 
zontal chords  of  the  curve  of  shade,  one  on  each  nappe  of  the 
piedouche. 

d.  It  may  be  shown  in  several  ways,  that  there  must  be  such 
points  of  maximum  width  of  the  curve,  at  which  tangent  rays 
may  be  drawn  to  the  curve.     This  may  be  done,  in  an  elementary 
manner,  as  follows.     Suppose  all  points  of  the  curve  of  shade  to 
be  found  by  the  general  method  of  (74)  employed  in  finding  the 
highest  and  lowest  points  of  shade.     A  vertical  plane  of  rays 
near  the  gorge,  will  cut  a  curve  similar  to  A"B",  PI.  X.,  Fig.  32, 
to  which  two  tangent  ra}rs  can  be  drawn,  giving  two  points  of 
shade,  T  and  T'.     Consideration  of  the  form  of  the  piedouche 
will  show,  that,  as  the  vertical  secant  plane  recedes  from  the 
gorge,  the  curve,  A"B",  will  become  flatter,  the  points  T  and  T' 
will  approach  each  other,  and  will  finally  unite,  as  at  T,  on  A'B'; 
T  being  a  point  of  inflexion,  or  change  of  curvature. 

If  the  cutting  plane  recedes  still  further  from  the  gorge,  the 
curve  will  be,  as  at  AB,  so  flat  that  no  tangent  ray  can  be  drawn 
to  it ;  hence  the  point  T  marks  the  position  of  the  greatest  width 
of  the  curve. 

e.  Moreover,  the  ray  is  tangent,  at  T,  to  the  curve  of  shade 
itself,  for,  on  that  position  of  A"B"  which  is  consecutive  with 


SHADES  AND  SHADOWS.  81 

A'B',  T  and  T'  will  be  consecutive  points  of  shade,  and  their 
tangents  consecutive  parallels.  On  A'B',  therefore,  T  and  T' 
unite,  and  their  separate  tangents  coalesce,  forming  a  tangent  to 
the  curve  of  shade  at  T. 

f.  Those  of  the  secant  planes  which  intersect  the  gorge,  cut 
the  piedouche  in  curves  analogous  to  the  meridian  curve. 

g.  Furthermore,  the  points,  as  T,  Fig.  32,  near  1,  2,  E,  etc., 
on  PL  X.,  Fig.  31,  are  the  limits  between  that  part  of  the  curve 
of  shade  which  really  exists  on  the  exterior  of  a  solid  opaque 
piedouche,  and  on  a  transparent  one,  open  at  the  top,  and  con- 
sisting merely  of  a  surface  without  thickness.     All  points,  as  T, 
on  the  upper  nappe,  and  similar  ones  at  such  points  at  T',  on  the 
curves  like  AX/B7/  cut  from  the  lower  nappe,  are  imaginary  points, 
on  the  interior  side  of  the  hollow  surface.     Only  such  points  as 
T',  on  curves  shaped  like  A^B"  and  cut  from  the  upper  nappe, 
and  such  as  T,  situated  on  curves  cut  from  the  lower  nappe,  are 
points  in  which  rays  in  space  are  truly  tangent  to  the  external 
surface  of  the  piedouche.     The  construction  of  a  few  intersec- 
tions of  the  piedouche  by  vertical  planes  of  rays  will  make  these 
statements  clear. 

h.  There  is  a  direct  construction,  given  by  M.  LEROY,  of  the 
points,  as  T,  of  maximum  width  of  the  curve  of  shade,  but  it  is 
extremely  tedious  and  comparatively  unimportant,  being,  in  the 
graphical  point  of  view,  only  approximate.  Hence  it  is  not 
here  given.  Moreover,  it  is  unimportant,  for  the  further  reason 
that  these  points  are  found  with  sufficient  accuracy  in  sketching 
the  curve  by  hand  through  a  considerable  number  of  other  con- 
structed points. 


§  IL — Shadows. 

82.  The  direct  method  of  finding  shadows  on  double  curved 
surfaces,  is  essentially  the  same  as  for  rinding  shadows  on  other 
surfaces.  Pass  a  plane  of  rays  through  any  point,  P,  of  the  line 
casting  ike  shadow,  and  construct  the  intersection,  I,  of  this  plane 
with  the  given  double  curved  surface.  The  point,  where  a  ray  through 
P  meets  the  line,  I,  is  the  shadow  of  P  on  the  double  curved  surface. 

It  being  desirable,  for  graphical  convenience,  that  the  line  I 
should  be  a  circle,  it  is  evident  that  the  foregoing  direct  method 
is  always  conveniently  applicable  only  to  spheres ;  and  to  other 

6 


82  GENERAL  PROBLEMS. 

surfaces  of  revolution,  only  when  so  situated  in  reference  to  the 
light,  that  planes  of  rays  may  be  made  to  cut  them  in  circles. 
As  these  conditions  do  not  generally  occur,  the  shadows  on 
double  curved  surfaces  are  mostly  found  by  a  variety  of  indirect 
and  special  methods,  which  will  be  sufficiently  understood  from 
the  following  problems. 


PROBLEM  XXIX. 

To  construct  tJie  shadow  of  the  front  circle  of  a  niche  upon  its  spheri- 
cal part. 

Principles. — This  problem,  enunciated  more  as  an  abstract  one, 
is  this :  To  construct  the  shadow  of  a  vertical  circle  upon  the 
interior  of  a  hemispherical  surface  of  which  the  given  circle  is 
the  front  edge. 

83.  If  a  straight  line  be  moved,  so  as  to  remain  constantly 
parallel  to  its  first  position,  and  to  rest  on  the  circumferences  of 
two  great  circles  of  a  sphere,  it  will  generate  a  cylinder,  of  which 
those  circles  will  be  equal  oblique  sections.  All  the  elements  of 
this  cylinder  will  intersect  the  sphere,  except  the  two  which  are 
tangent  to  it  at  the  extremities  of  the  common  diameter  of  the 
two  great  circles.  Hence  we  have  the  principle,  that  a  cylinder 
which  intersects  a  sphere  in  a  great  circle,  intersects  it  also  in  a  second 
great  circle.  The  right  section  of  this  cylinder  will  be  an  ellipse, 
equally  inclined  to  these  great  circles,  and  whose  transverse  axis 
is  their  common  diameter. 

It  now  follows,  that  the  cylinder  of  rays  passing  through  the 
front  edge  of  the  quarter  sphere,  will  intersect  its  interior  in  an 
arc  of  a  great  circle  of  that  sphere,  which  arc  will  be  the  shadow 
required. 

1°.  Tlie  first  solution,  based  on  the  foregoing  general  princi- 
ples, is  made  by  the  direct  general  method  (82),  and  depends  imme- 
diately on  the  following — 

Principles. — A  plane  of  rays,  perpendicular  to  the  vertical  plane 
of  projection,  will  cut  a  circular  arc  from  the  spherical  part  of  the 
niche,  and  a  ray  from  the  cylinder  of  rays,  whose  intersection  will 
be  a  point  of  the  required  shadow. 

Construction. — PI.  X.,  Fig.  33.  In  this  construction,  three 
planes  of  projection  are  used. 


SHADES  AND   SHADOWS.  83 

The  plane  of  the  front  of  the  niche  is  taken  as  the  vertical  plane 
of  projection.  The  plane  of  the  upper  base  of  the  niche  is  taken 
as  the  horizontal  plane  of  projection,  AC  is  therefore  the  ground 
line  for  these  planes,  and  the  spherical  quadrant  of  the  niche  is 
in  their  second  angle.  D'H',  parallel  to  the  vertical  projection, 
DO,  of  a  ray  of  light,  is  the  trace,  or  ground  line,  of  the  auxili- 
ary plane  of  projection.  D'H'  is  in  the  vertical  plane,  and  the 
auxiliary  plane  is  revolved  about  it,  into  the  plane  of  the  paper, 
so  that  the  part  in  front  of  the  vertical  plane  falls  to  the  right 
of  D'H'. 

DO — D"p  is  a  ray  of  light.  When  a  point  is  projected  on 
two  planes,  each  perpendicular  to  a  third,  its  projections  on  those 
planes  are  at  equal  distances  from  their  respective  ground  lines, 
hence,  making  O'p'  equal  to  Op,  pf  is  the  auxiliary  projection 
of  the  point  O^p  of  the  ray  DO — D"p.  D  is  projected  at  D', 
therefore  D'p'  is  the  projection  of  the  ray,  DO — D'^9,  on  the 
auxiliary  plane. 

The  plane  of  rays,  whose  vertical  trace  is  DO,  cuts  from  the 
spherical  part  of  the  niche,  produced,  the  semicircle  DOc — • 
D'T'H'  and  from  the  cylinder  of  rays  (83),  the  ray,  DO— D'/, 
which  pierces  the  semicircle  at  c',  which,  being  projected  back 
at  c,  gives  c,cf  as  one  point  of  the  required  shadow — produced. 

Likewise,  any  other  parallel  plane  of  rays,  as  ak,  cuts  from 
the  niche  the  semicircle  ak — a'h'f",  and  from  the  cylinder  of 
rays,  the  ray  ah — a'h',  giving  h,hf  for  another  point  of  shadow. 

As  N'T'  is  the  ray  which  is  the  tangent  element  (83)  of  the 
cylinder  of  rays,  T'  is  where  the  shadow  begins,  and  T7ic  is  the 
quadrant  of  shadow,  on  the  spherical  part  of  the  niche,  pro- 
duced. 

2°.  The  second  Solution  is  made  by  the  special  method  of  (42). 

Construction.— PI.  X.,  Fig.  34,  CHB— C'A'B'  is  the  upper 
base  of  the  semi-cylindrical  part  of  the  niche.  CAB — C'A^B' 
is  the  front  edge  of  its  spherical  part. 

By  the  first  solution,  T,Tr,  where  a  plane  of  rays,  N'T',  perpen- 
dicular to  the  vertical  plane  of  projection,  is  tangent  to  the 
spherical  part,  on  its  front  edge,  is  the  point  where  the  required 
shadow  on  the  spherical  interior  begins. 

To  find  another  point,  take  any  auxiliary  plane,  parallel  to  the 
front  face  of  the  niche.  EH  may  be  assumed  as  the  trace  of 
such  a  plane,  on  the  plane  C'A'B'.  The  ray  AE— A'E', 
through  the  centre  of  the  front  semicircle,  pierces  the  vertical 


84  GENERAL   PROBLEMS. 

plane  EH,  at  E,E',  which  is  therefore  the  centre  of  the  shadow 
of  that  semicircle  on  the  plane  EH.  With  E'  as  a  centre,  and 
a  radius  equal  to  A'C'  (33),  describe  the  arc  which  intersects  the 
semicircle  Wh',  cut  from  the  quarter  sphere  by  the  plane  EH, 
at  h' ;  whose  horizontal  projection  is  7i,  on  the  trace  EH.  hfi  is 
therefore  a  required  point  of  shadow. 

The  point  y,yf  is  similarly  found,  on  a  similar  vertical  secant 
plane,  aY. 

3°.  Construction  of  the  point  where  the  shadow  leaves  the  sphwical 
part  of  the  niche. 

This  point,  Z>,5',  may  here  be  found,  without  reference  to  the 
cylindrical  part  of  the  niche. 

Remembering  that  the  plane  of  shadow  contains  a  great  circle 
of  the  spherical  part,  the  diameter  A'T'  is  its  trace  on  the  front 
face  of  the  niche,  and  therefore  A' A  is  one  point  of  its  trace  on 
the  base,  CHB,  of  the  spherical  part.  That  trace  intersects  the 
semicircle  CHB  at  the  point  required.  Here  then  is  given  one 
trace,  A'T',  of  a  plane,  and  a  point,  as  y,y',  in  that  plane,  to  find 
its  other  trace.  Now  ya — y'a'  is  a  line  through  y,y',  and  paral- 
lel to  the  trace  A'T',  and  therefore  is  a  line  of  the  plane  of 
shadow.  This  line  pierces  the  plane  C'A'B'  at  a',a,  giving  Aab 
for  the  required  trace  on  C'A'B',  and  b,b'  for  the  required  point 
at  which  the  shadow  leaves  the  spherical  part  of  the  niche ;  it 
being  the  intersection  of  the  plane  of  shadow  with  the  base 
line  CHB. 

4°.  Third  Solution. — This  is  made  by  the  special  method  of 
(43)  which,  in  its  present  application,  depends  immediately  on 
the  following — 

Principles. — Knowing,  in  advance,  from  (83)  that  the  curve 
of  shadow  is  a  plane  curve,  we  can  find  its  rectilinear  projection 
on  a  plane  of  projection,  taken  perpendicular  to  the  plane  of 
shadow.  Then,  having  this  rectilinear  projection,  both  projec- 
tions of  particular  points  in  it  can  be  found  as  in  (43). 

Construction. — PI.  X.,  Fig.  33.  D'H'  is  the  trace,  on  the  plane 
of  the  face  of  the  niche,  of  a  plane  perpendicular  to  the  plane 
of  shadow.  All  the  planes  of  projection  are,  moreover,  the 
same  here,  as  in  the  first  solution. 

Having  found,  as  before,  c',  we  have  OT'Y  for  the  vertical, 
and  O'c'  for  the  auxiliary,  trace  of  the  plane  of  the  shadow,  the 
latter  trace  being  also  the  projection  of  the  shadow  on  the  auxili- 
ary plane.  Any  semicircle,  as  a'h'f",  lying  in  the  spherical 


SHADES  AND  SHADOWS.  85 

surface,  or  any  ray,  as  a/A',  will  cut  from  the  given  projection  of 
the  shadow  a  point,  as  A',  whose  other  projection,  A,  is  found  by 
drawing  h'h,  perpendicular  to  the  ground  line  D'H',  and  noting 
its  intersection,  A,  with  the  vertical  projection,  ok,  of  the  same 
semicircle,  a'A/",  or  ray  a'h'. 

Discussion. 

FIRST. — If  it  were  not  known  already  that  the  shadow  on  the 
spherical  part  of  the  niche  is  a  plane  curve,  it  could  be  proved  by 
reference  to  PI.  X.,  Fig.  33.  For,  O'D'  and  O'c'  are  equal,  hence 
O'c'D'  is  an  isosceles  triangle.  But  a'h'  being  a  chord,  parallel 
to  D'c',  in  a  semicircle  a'h'f",  concentric  with  D'c'H',  it  follows 
that  OVA'  is  an  isosceles  triangle  also,  and  is  similar  to  O'D'c'. 
Then,  as  0V  and  O'D'  coincide  in  direction,  O'A'  and  O'c'  like- 
wise coincide,  and  O'A'c',  the  projection  of  the  shadow  on  the 
auxiliary  plane,  is  a  straight  line,  which  shows  the  shadow  itself 
to  be  a  plane  curve. 

SECOND. — Another  construction  of  the  point,  e,  where  the  sha- 
dow leaves  the  spherical  part  of  the  niche,  may  here  be  given. 
C'B"  is  the  auxiliary  projection  of  a  line,  perpendicular  to  the  ver- 
tical plane  of  projection  at  C.  Its  horizontal  projection  would  be 
a  perpendicular  to  the  ground  line,  AC,  at  C  ;  but,  to  avoid  con- 
fusion of  the  figure,  a  similar  line  is  shown  at  AB.  Producing 
O'c'  to  B",  gives  B"  as  the  point  in  which  the  plane  of  shadow 
intersects  C'B".  Now  makeAB  equal  to  C'B",  and  OB  will  be 
the  horizontal  trace  of  the  plane  of  shadow,  after  revolving  180° 
about  OU,  in  the  horizontal  plane,  as  an  axis.  The  point  m  is 
then  the  revolved  position  of  the  intersection  of  the  horizontal 
trace  of  the  plane  of  shadow,  with  the  horizontal  semicircle, 
AUC,  of  the  niche.  This  intersection  is  the  point  sought.  Then 
the  horizontal .  projection,  ran,  of  the  vertical  arc  of  counter- 
revolution, gives  n  for  the  primitive  horizontal  projection  of  this 
point,  and  ne,  perpendicular  to  the  ground  line,  AC,  gives  e,  the 
vertical  projection  of  the  same  required  point  of  shadow. 

THIRD. — The  point  e  can  again  be  found,  but  considered  still 
as  the  intersection  of  the  horizontal  trace  of  the  plane  of  shadow 
with  the  horizontal  semicircle,  AUC,  of  the  niche.  This  is  done 
by  viewing  this  trace  as  the  intersection  of  the  plane  of  shadow 
with  the  plane  of  the  horizontal  circle  AUC.  Then,  after  show- 
ing the  traces  of  both  of  these  planes  on  the  planes  of  projection 


86  GENEEAL   PEOBLEMS. 

•whose  ground  line  is  D'H',  their  line  of  intersection  can  be 
found. 

The  traces  of  the  plane  of  shadow  are  OT'Y,  on  the  vertical 
plane,  and  O'B",  on  the  auxiliary  plane  D'H'.  The  traces  of  the 
plane  of  the  base  of  the  niche  are  T'O' Y,  on  the  auxiliary  plane, 
and  O'e",  on  the  vertical  plane ;  for,  thus  considered,  D'H'  is 
supposed  to  be  the  trace  of  the  plane  DO  itself,  after  being  trans- 
lated, parallel  to  itself,  to  the  position  D'H',  and  then  revolved, 
as  before  described.  O'e"  is  thus  parallel  to  OC,  being  only  a 
transferred  position  of  OC. 

The  intersection  of  the  plane  of  shadow  and  the  plane  of  the 
base  is  now  projected  in  O'c'  and  O'e".  The  arc/"/"  represents 
a  revolution  of  the  plane  of  the  base,  together  with  the  intersec- 
tion just  noted,  about  O'— T'O'Y  as  an  axis,  and  into  the  trans- 
lated position,  D'H',  of  the  plane  of  rays  DO.  The  base  then 
appears  in  D'c'II',  and  its  intersection,  O'e" — O'c',  with  the 
plane  of  shadow,  at  O'f.  At  e',ef"  is  therefore  the  revolved 
position  of  the  point  of  shadow  on  the  base  of  the  niche.  In 
counter-revolution,  e'e"  returns  to  e"E,  from  which  is  projected, 
at  e,  the  required  projection  of  the  required  point  of  shadow.  Or, 
e'  may  be  first  counter-projected  at  e'"' ',  and  then  counter-revolved, 
as  shown  by  the  arc  e""e,  which  gives,  again,  the  same  required 
point,  e. 

FOURTH. — Considering  D'H',  again,  as  the  ground  line  of  the 
original  position  of  the  auxiliary  plane  of  projection,  e  may  again 
be  found  by  the  intersection  of  the  projection,  O'c',  of  the  shadow 
on  the  spherical  quadrant,  with  the  auxiliary  projection,  C'ET', 
of  the  base,  AUC,  of  the  niche.  C'ET'  is  a  quarter  ellipse  on 
the  semi-axes  O'T'  and  O'C',  C'  being  the  projection  of  C. 

FIFTH. — A  little  consideration  will  show  that  C'O'  :  C'H':: 
Er  I  Ee'. 

Hence  O'T',  C'E,  and  H'e'  all  meet  at  I.  Hence,  from  any 
point,  I,  on  O'O,  draw  lines  to  C'  and  H'.  From  the  intersec- 
tion, as  e',  of  IH'  with  the  semicircle  D'c'IF,  draw  a  perpendicu- 
lar to  O'O  ;  its  intersection,  as  E,  with  1C'  will  be  a  point  of  the 
ellipse. 

Thus,  after  finding  e',  as  above,  we  may  find  E,  and  hence  e, 
by  drawing  H'e'I,  and  then  1C'  which  will  intersect  O'c'  at  E  ; 
•which  is  projected  by  E<?,  perpendicular  to  D'H'  at  e. 

Moreover,  we  have  here  a  construction  of  the  ellipse  by  homo- 
logous secants,  analogous  to  the  homologous  tangents  which  may 


SHADES  AND  SHADOWS.  87 

be  drawn  at  e'  and  E,  and  which  will  meet  O'O  at  a  common 
point. 

SIXTH. — Finally,  e  may  be  approximately  found  as  the  point 
where  the  entire  quadrant  of  shadow,  The,  on  the  spherical  part 
produced,  crosses  AC.  (See  42.) 

SEVENTH. — To  construct  the  tangent  line  at  the  point,  e. 

Principles. — At  the  point  e,  the  required  line  is  tangent  both  to 
the  spherical  branch,  ehT',  and  to  the  cylindrical  branch  of  the 
shadow.  For,  the  tangent,  L,  to  the  spherical  branch,  is  the 
intersection  of  a  tangent  plane,  P,  to  the  sphere  of  the  niche  at  <?, 
with  the  tangent  plane,  P',  to  the  cylinder  of  rays,  along  the  ele- 
ment eW.  The  tangent,  L',  to  the  cylindrical  branch,  is  the 
intersection  of  the  same  tangent  plane,  P',  to  the  cylinder  of  rays, 
along  the  element  eE',  with  the  tangent  plane,  P",  along  the 
element,  eB",  of  the  cylinder  of  the  niche.  Now,  since  this 
cylinder  and  the  sphere  are  tangent  to  each  other  on  AUC,  the 
planes,  P  and  P",  coincide,  hence  the  tangent  lines,  L  and  L', 
coincide  also,  forming  a  common  tangent  to  the  two  curves  of 
shadow  which  coalesce  at  e. 

Again  :  The  curve  ehT'  is  a  plane  curve,  and  the  tangent  to 
a  plane  curve  of  intersection  is  the  intersection  of  the  plane  of 
that  curve,  with  a  tangent  plane  to  the  surface  on  which  it.  lies. 
Hence,  finally,  the  required  tangent  line  is  best  found  as  the 
intersection  of  the  tangent  plane  to  the  cylinder  of  rays,  along  the 
element  eE,  with  the  plane  of  the  curve  of  shadow. 

Construction. — PI.  X.,  Fig.  33.  E'Y,  tangent  to  the  front  of 
the  niche,  or  base  of  the  cylinder  of  rays,  at  E',  is  the  vertical 
trace  of  the  tangent  plane  to  the  cylinder  of  rays,  along  the  ele- 
ment eFj'.  OT'Y  is  the  vertical  trace  of  the  plane  of  shadow. 
Then  Ye,  joining  the  intersection  of  these  traces  with  the  given 
point  e,  which  is  common  to  both  planes,  is  their  intersection,  or, 
the  required  tangent  line. 

EIGHTH. — Having  now,  by  previous  constructions,  the  tangents 
at  T',  b',  and  c,  PI.  X.,  Fig.  34,  where  c  represents  the  upper  end 
of  the  vertical  shadow  of  D'C',  it  is  easy  to  construct,  according 
to  elementary  plane  geometry,  the  curved  shadow  by  approxi- 
mate circular  arcs.  Thus,  from  T'  to  V  may  be  made  part  of  an 
oval  of  three  centres,  having  AT'  and  A'E'  for  its  axes ;  and  b'c 
may  be  a  single  arc,  tangent  to  the  tangents  at  b'  and  c,  or,  if 
necessary,  it  may  be  an  arc  of  a  compound  curve  having  the  same 
tangents,  and  a  horizontal  line  at  c  for  one  axis.  When  the 


88  GENERAL   PROBLEMS. 

projections  of  the  light  make  an  angle  of  45°  with  the  ground 
line,  a  single  arc  can  be  drawn  through  b'  and  tangent  at  T',  and 
another,  through  b'  and  tangent  at  c;  which  will  nearly  coincide 
with  the  shadow. 


PROBLEM  XXX. 

{To  find   the  shadow  of  the  upper  circle  of  a  piedouche  upon  its 

concave  surface. 

Principles. — The  highest  and  lowest  points  of  this  shadow  are 
found  by  the  direct  general  method  (41).  Other  points  are  found 
by  the  special  method  of  (42). 

Construction. — PL  XL,  Fig.  35,  shows  the  surface  represented 
as  in  PI.  X..,  Fig.  31.  Its  projections  need  not  again  be  described. 

1°.  The  highest  and  lowest  points  of  shadow. 

AK  is  the  horizontal  trace  of  a  meridian  plane  of  rays.  Since 
the  piedouche  is  a  surface  of  revolution,  if  we  revolve  AK  about 
the  vertical  axis  at  A,  till  it  becomes  parallel  to  the  vertical  plane 
of  projection,  the  meridian  curves  contained  in  it  will  coincide 
with. E'c^"  and  B'D'C',  which  are  seen  in  the  vertical  projection, 
of  the  piedouche. 

F,F'  is  the  revolved  position  of  the  point  whose  horizontal 
projection  is  F",  and  which  is  cut  by  the  plane  of  rays  from  the 
upper  circle  casting  the  shadow.  Then  the  revolved  position  of 
a  ray  through  F,F',  will  intersect  F'cT'E'  and  B'D'C',  respec- 
tively, in  the  revolved  positions  of  the  highest  and  lowest  points 
of  shadow,  viz.  at  d'",d'r,  and  r"V",  in  the  meridian  plane  FAC. 

The  revolved  ray,  F  W'V",  is  parallel  to  AV",  which  is  found 
by  revolving  the  ray  Ac — AV  to  the  position  Ac" — AV", 
parallel  to  the  vertical  plane  of  projection. 

In  the  counter-revolution,  d",d"'  and  r",r'"  return,  respectively, 
in  the  horizontal  arcs,  d"d—d'"d',  and  r'"r — r"'r't  to  their  true 
positions,  at  d,d'  and  r"V,  as  the  highest  and  lowest  points  of 
the  required  shadow. 

2°.  To  find  any  intermediate  points. 

Any  horizontal  plane,  as  j'D',  cuts  from  the  surface  of  the 
piedouche  a  circle,  q'D' — qsf;  and  from  the  cylinder  of  rays, 
whose  given  base  is  the  circle,  BNF — F'A'B',  a  circle  equal  to 
the  latter  circle,  and  which  is  the  shadow  of  BNF— F'A'B'  on 


SHADES  AND  SHADOWS.  89 

ihe  plane  ql)'.  The  intersection  of  this  auxiliary  shadow  with 
the  circle  qsf—q'D',  gives  two  points  of  shadow  on  the  given 
surface  (42). 

The  centre  of  the  auxiliary  shadow  is  b\b-  where  the  ray  Ab — 
A'6',  through  the  centre  of  the  upper  circle  of  the  concave  sur- 
face, pierces  the  plane  g-T)'.  Then,  drawing  an  arc,  fq,  with  a 
centre,  &,  and  a  radius  equal  to  A'B',  we  find /and  q,  horizontal 
projections  of  two  points  of  the  required  shadow.  Projecting 
these  points  on  q'T>',  the  vertical  trace  of  the  horizontal  plane 
containing  them,  gives  q'  and/  for  their  vertical  projections. 

Two  points  can  be  similarly  found,  on  any  other  auxiliary 
horizontal  plane  between  the  highest  and  lowest  points,  as  shown 
at  e,e',  on  the  plane  e'a',  and  &tp,pf  on  the  plane  p'c' . 

Discussion. 

FIRST. — The  whole  of  the  shadow  just  found,  would  be  real, 
only  in  a  geometrical  sense ;  or  upon  a  hollow  and  transparent 
piedouche.  Observing  that  the  highest  point  of  the  curve  of  shade 
would  be  found  by  drawing  a  tangent  ray,  parallel  to  A'c'",  and 
counter- revolving  as  before,  it  is  plain  that  the  highest  point  of 
shadow  is  below  the  highest  point  of  shade.  Hence  the  shadow 
on  an  opaque  solid  piedouche  will  be  real,  from  its  highest 
point  to  its  intersection  with  the  curve  of  shade.  The  real  part 
of  the  curve  of  shade  was  described  in  Prob.  XXVIII.  (Eem.^). 
This  real  part  will  cast  a  shadow  on  the  lower  part  of  the 
piedouche. 

Thus  the  complete  line  of  separation,  between  the  illuminated 
and  the  dark  portions  of  the  piedouche,  consists  of  the  real  por- 
tion of  the  shadow  of  the  upper  circle,  the  real  part  of  the  curve 
of  shade,  and  the  shadow  of  this  real  shade. 

SECOND. — The  latter  shadow  is  thus  found.  See  PI.  XI.,  Fig. 
36,  which  represents,  in  horizontal  projection,  and  sufficiently  for 
purposes  of  explanation,  the  shadow  just  mentioned,  together 
with  the  entire  shadow  of  the  piedouche  on  the  horizontal 
plane. 

The  shadow  of  the  curve  of  shade  upon  the  lower  part  of  the 
piedouche,  is  found  by  the  special  method  of  two  auxiliary 
intersecting  shadows  (66-ls/.  Eem.  a).  Thus,  qkt  and  pmu  are 
shadows  of  the  real  parts  of  the  curve  of  shade,  on  the  horizontal 
plane.  They  are  found  from  the  projections  of  those  parts, 


90  GENERAL   PEOBLEMS. 

shown  in  PI.  X.,  Fig.  31,  and  just  as  any  shadow  is  found  on  a 
plane  of  projection  (30). 

The  circle  sr  is  the  shadow  of  an  equal  circle,  ITT,  near  the 
foot  of  the  piedouche ;  r  and  s,  the  intersections  of  these  sha- 
dows, are  the  shadows  of  points  of  shadow  cast  by  the  curve 
of  shade  on  the  circle  UT.  That  is,  rays,  as  sU  and  rT,  through 
5  and  r,  will  meet  both  UT,  and  the  curves  of  shade.  U  and  T 
are  the  shadows,  on  UT,  of  the  points  cut  from  the  curve  of 
shade  by  these  rays. 

THIRD. — pq,  and  ut,  are  the  shadows  of  the  unreal  portions, 
1U2,  and  QrE,  PL  X..  Fig.  31,  of  the  curve  of  shade. 

These  unreal  shadows  join  the  real  ones  at  p,  q,  u,  and  t, 
forming  cusps  of  the  first  order. 

This  result  being  apparent,  it  enables  us  to  learn,  as  we  could 
not  easily  do  from  an  inspection  of  the  curve  of  shade  only,  in 
PI.  X.,  Fig.  31,  the  real  nature  of  the  tangent  cylinder  of  rays, 
whose  curve  of  contact  with  the  piedouche  is  the  curve  of 
shade. 

Considering  putq  as  its  base,  and  that  its  elements  are  parallel, 
it  is  now  evident,  that  above  the  point  2,2',  for  example,  its  sur- 
face curves  upward,  and  backward  towards  the  vertical  plane ; 
and  that  below  the  same  point,  its  surface  bends  downward,  and 
backward  towards  the  vertical  plane ;  so  that  it  has  an  edge  on 
the  tangent  ray  at  2,2'.  A  plane  through  this  ray,  or  element, 
and  tangent  on  the  same,  to  both  the  branches,  j  ust  described, 
of  the  cylinder,  is  an  osculatory  plane  to  the  curve  of  shade  at 
2,2' ;  and  its  horizontal  trace  is  a  tangent  to  the  base  of  the 
cylinder  at  q,  PL  XL,  Fig.  36,  between  hq  and  pq.  But  the 
tangent  ray  at  2,2',  being  such  a  salient  edge  as  just  described,  a 
plane  section,  as  the  base  of  the  cylinder,  will  present  a  cusp  as 
at  q,  where  that  edge  pierces  the  plane  of  the  section. 

Observing,  further,  that  the  cylinder  of  rays,  to  which  the 
plane  of  rays  is  tangent,  is,  itself,  tangent  to  the  piedouche  at  R, 
the  plane  is  also  tangent  to  the  piedouche  at  R.  Hence  its  hori- 
zontal trace  can  readily  be  found. 

FOURTH.— The  rest  of  the  shadow,  PL  XI.,  Fig.  36,  is  thus 
composed :  ahc  is  the  shadow  of  that  lower  semicircle  of  the 
upper  base  of  the  piedouche  which  is  towards  the  light,  deb  is 
the  shadow  of  the  opposite  upper  semicircle,  KSB.  ab  and  cd 
are  the  shadows  of  the  elements  of  shade  of  the  upper  base  at  K 
and  B.  Likewise,  nmy  is  the  shadow  of  the  upper  semicircle, 


SHADES   AND   SHADOWS.  91 

NHY,  of  the  lower  base  of  the  piedouche,  and  ~Nn  and  Yy  are 
the  shadows  of  the  elements  of  shade  of  the  lower  base,  at  N 
and  Y 


§  III. — General     Problem    in    Review    of    Shades    and 
Shadows,  determined  by  Parallel  Rays. 

84.  No  better  problem  can  perhaps  be  found  for  this  review 
than  that  of  the  shades  and  shadows  on  the  Roman  Doric  column, 
embracing,  as  it  does,  the  shadows  both  of  straight  and  of  curved 
lines ;  on  planes,  and  on  a  variety  of  single  curved,  and  double 
curved  surfaces. 

Description  of  the  Column. — PI.  XL,  Fig.  37,  represents,  with- 
out regard  to  precise  architectural  proportions,  'a  fragment  of  the 
shaft  and  base  of  the  column. 

Indicating  its  horizontal  circles  by  their  radii,  the  circle  Ca  is 
the  plan  of  the  shaft  a'SK.  Next,  Cb  is  the  plan  of  the  smallest 
section  of  the  scotia  e'b'c".  Then  Cc  is  the  plan  of  the  upper 
and  middle  fillets,  c'g'  and  c"h'.  And  Cd  is  the  largest  section  of 
the  upper  torus,  c"d'h'g'.  Ce — e'n'  is  the  lower  fillet.  CF — -f'nf  is 
the  lower  torus.  FGH — F'H/  is  a  square  member,  called  the 
plinth,  a! eg'  is  the  foot  of  the  shaft,  and  is  a  concave  double 
curved  surface.  The  fillets  are  short  vertical  cylinders. 

PI.  XL,  Fig.  38,  similarly  represents  the  principal  parts  of  the 
capital  and  shaft  of  the  column.  a"ahw — a'b'io'g"h'  is  the 
abacus,  whose  horizontal  sections  are  squares.  The  portion, 
on — o'n'n",  of  the  abacus,  is  the  cyma  reversa,  and  is  cylindrical. 
Cg  and  C/are  the  greatest  and  least  circular  sections  of  the  half 
torus  g'g'f'i  called  the  echinus.  Cf—f'f"  is  a  fillet.  Between 
C/and  Ge  is  e'f'f ',  the  cavetto.  Cc  is  the  neck  of  the  column, 
limited  by  a  small  torus,  not  shown,  and  called  the  astragal,  just 
under  which  is  the  lower  fillet  and  then  a  double  curved  concave 
surface  similar  to  the  foot  of  the  column. 

Thus  the  whole  column,  between  the  plinth  and  the  abacus,  is 
a  surface  of  revolution  having  a  vertical  axis. 


92  GENERAL  PROBLEMS. 


PROBLEM  XXXI. 

To  find  the  Shades  and  Shadows  of  the  Shaft  and  Base  of  a  RomLn 
Doric  Column. 

Such  parts  of  the  solution  as  present  peculiar  features,  will  be 
figured.  Others  will  be  merely  referred  to  previous  problems 
containing  similar  constructions. 

1°.  The  shaft  is  strictly  a  double  curved  surface,  its  upper 
diameter  being,  in  practice,  a  little  less  than  its  lower,  and  its 
generatrix  slightly  curved.  The  fragments  shown  in  the  figures 
may,  however,  be  treated  as  vertical  cylinders,  in  finding  their 
elements  of  shade. 

2°.  The  shadow  of  the  shaft  on  the  foot  of  the  column,  is 
found  by  the  special  method  of  (43)  (Prob.  XVL,  3°),  its  hori- 
zontal projection  being  known  as  a  straight  line,  tangent  to  the 
plan  of  the  shaft. 

3°.  The  element  of  shade,  and  the  upper  circle  of  the  fillet,  cast 
shadows  on  the  upper  torus.  See  PL  XII.,  Fig.  39. 

The  element  of  shade  is  the  vertical  line  at  NN7,  and  its  shadow 
is  found  as  in  (2°).  Thus,  NK  is  the  indefinite  horizontal  pro- 
jection of  the  shadow,  NK  being  the  intersection  of  a  vertical 
tangent  plane  of  rays,  at  N",  with  the  torus,  and  therefore  paral- 
lel to  the  horizontal  projections  of  rays,  da — d'a!  is  any  assumed 
horizontal  circle  of  the  torus,  intersecting  the  indefinite  shadow 
NK  at  a!  whose  vertical  projection,  a',  is  on  d'a',  the  vertical 
projection  of  the  assumed  circle. 

This  shadow  begins  at  N',  the  foot  of  the  element  of  shade. 

The  upper  circle  casts  a  shadow  which  is  found  by  the  method 
of  (42),  see  also  (Prob.  XXX— 2°).  Thus,  the  ray,  G/i—G'h't 
through  the  centre  of  the  upper  circle,  pierces  an  auxiliary  hori- 
zontal plane,  h'e',  cutting  the  torus  in  a  circle,  at  h,hf.  The  cir- 
cle, ybj  with  centre  h  and  radius  hy,  equal  to  GV,  is  the  auxiliary 
shadow  of  the  upper  circle,  GV,  on  the  plane  h'e'.  This  shadow 
meets  the  circle  eb,  cut  from  the  torus  by  the  plane  h'e',  at  two 
points,  one  of  which  is  b.  Then  b  is  vertically  projected  at  &', 
on  the  vertical  projection,  e'k',  of  the  circle  cut  from  the  torus. 

On  a  complete  construction,  the  latter  shadow  will  intersect 
the  shadow  of  the  element  of  shade,  just  before  found,  and  the 


SHADES  AND  SHADOWS.  93 

curve  of  shade  of  the  upper  torus,  in  points  which  the  student 
may  construct. 

4°.  The  curve  of  shade  on  each  torus  in  found  as  in  Prob. 
XXVII. 

5°.  The  curve  of  shade  of  the  upper  torus  casts  a  shadow  on 
the  middle  fillet,  which  is  found  as  in  Prob.  IV.  But  PI.  XIL, 
Fig.  39,  serves,  however,  to  show  how  this  shadow  can  be  found, 
having  given  but  one  projection  of  the  curve  of  shade  of  the 
torus.  Let  EV  be  a  fragment  of  the  vertical  projection  of  the 
curve  of  shade  of  the  torus.  Let  o'  be  a  point  on  this  curve, 
which  is  supposed  to  cast  a  shadow  on  the  fillet.  This  point  is 
transferred,  on  a  circular  section  of  the  torus,  to  a  point  whose 
revolved  position  is  D',D",  found  by  revolving  the  semicircular 
meridian  curve,  in  the  plane  DGGr',  about  its  vertical  diameter 
atS, 

In  the  counter  revolution  about  S,  D'D"  proceeds  to  D,  in 
horizontal  projection  ;  and  in  the  counter  revolution  about  the 
vertical  axis  at  G,  it  returns  to  o,  the  desired  horizontal  projec- 
tion of  o'. 

This  done,  the  shadow  of  0,0'  is  found  by  the  simple  direct 
method  of  (40).  Thus,  op,  the  horizontal  projection  of  the  ray 
through  0,0',  pierces  the  fillet  at  a  point  of  shadow  whose  hori- 
zontal projection  is^>.  The  vertical  projection,  p',  of  this  point, 
is  at  the  intersection  of  the  projecting  line,  pp',  with  the  vertical 
projection,  o'p',  of  the  same  ray. 

6°.  Part  of  the  same  curve  of  shade  on  the  upper  torus,  casts 
a  shadow  on  the  scotia.  PI.  XIL,  Fig.  40.  This  curve  of  shade 
being  of  double  curvature,  its  shadow  on  an  auxiliary  plane,  P 
as  in  the  special  method  of  (42),  will  be  an  irregular  curve,  a  num- 
ber of  whose  points  must  be  found,  and  joined  together,  to  give 
the  auxiliary  shadow,  whose  intersection  with  the  circle  cut  from 
the  scotia  by  the  plane  P,  would  be  a  point  of  the  required 
shadow. 

Hence  the  special  method  of  (42),  which,  as  seen  in  Prob. 
XXX.,  is  so  convenient  when  the  shadow  is  cast  by  a  circle,  is 
no  more  convenient  than  the  direct  method  (82)  when  the  curve 
casting  the  shadow  is  of  double  curvature,  as  in  this  case. 

For  this  reason  the  direct  method  (S2)  as  applied  in  problems 
like  the  present,  is  here  used.  Let  b'c'  be  a  fragment  of  the  ver- 
tical projection  of  the  curve  of  shade  of  the  torus.  Let  V  be 
any  point  on  this  curve,  whose  shadow  on  the  scotia  is  to  be 


94  GENERAL   PROBLEMS. 

found.  It  is  transferred,  on  a  circular  section  of  the  torus,  to  a', 
whose  horizontal  projection  is  a.  In  counterrevolution  about  a 
vertical  axis  at  D ;  a',a  returns  to  b',b.  Now,  b'rf  is  the  vertical 
trace  of  a  plane  of  rays,  taken  through  &,&',  and  perpendicular 
to  the  vertical  plane  of  projection.  It  cuts  from  the  scotia  the 
curve  efr — e'f'r']  which  is  found,  as  fully  shown,  by  auxiliary 
horizontal  planes,  as  G'f.  This  plane  cuts  from  the  scotia  the 
circle  G'f — C/J,  and  from  the  curve  c'r'  the  point/',  whose  hori- 
zontal projection,  f,  is  on  the  horizontal  projection,  C/  of  the 
circle  containing  it.  Having  thus  constructed  the  intersection 
of  the  plane  of  rays  with  the  scotia,  by  means  of  a  system  of 
auxiliary  planes,  draw  the  ray  bs — b's',  which  pierces  the  curve 
efr — e'rf  at  5,5',  the  required  shadow  of  b,b'.  Other  points  of 
shadow  on  the  scotia  may  be  similarly  found. 

The  plane  of  rays  may  also  be  taken  vertical. 

7°.  There  will  be  two  points  of  the  curve  of  shade  b'c',  the 
rays  through  which  will  pass  through  the  points,  as  e,e',  on  the 
lower  edge  of  the  fillet.  That  part  of  the  curve  b'c',  lying 
between  these  points,  casts  a  shadow  on  the  fillet.  These  points 
themselves  cast  shadows  on  the  lower  edge  of  the  fillet,  and  the 
part  of  this  edge  between  these  points,  will  cast  a  shadow  on  the 
scotia.  The  latter  shadow  will  be  found  as  in  Prob.  XXX.  " 

8°.  The  curve  of  shade  of  the  scotia  will  be  found  as  in  Prob. 
XXVIIL 

9°.  If,  on  account  of  a  certain  direction  of  the  light,  the 
curve  of  shade  of  the  upper  torus  casts  a  shadow  on  the  lower 
torus,  it  will  be  indicated  by  the  passing  of  some  of  the  rays, 
similar  to  bs,  in  front  of  r,r',  so  as  not  to  cut  the  corresponding 
curve  similar  to  efr.  This  shadow  will  be  found  as  in  (6°). 

10°.  The  shadows  of  the  lower  fillet  on  the  lower  torus  are 
found  as  in  (3°). 

11°.  After  finding  the  curve  of  shade  on  the  lower  torus  as  in 
Prob.  XXVII.,  its  shadow  on  the  top  of  the  plinth  will  be  found, 
a  point  at  a  time,  as  in  Prob.  III. 


SHADES  AND  SHADOWS.  95 


PROBLEM  XXXII. 

To  find  Hie  Shades  and  Shadows  of  the  Capital  and  tihaft  of  a  Roman 
Doric  Column. 

In  making  the  solution  of  this  problem,  we  have — 

1°.  The  shadow  of  the  upper  member  of  the  abacus  on  the 
cyma  reversa ;  the  element  of  shade  on  the  latter ;  its  shadow 
on  itself,  and  on  the  lower  part  of  the  abacus.  These  shadows 
form  one  topic,  being  all  found  in  the  same  general  way,  as  seen 
in  PI.  XIL,  Fig.  41. 

The  cyma  reversa,  being  bounded  by  cylindrical  surfaces,  it  is 
necessary  to  have  an  auxiliary  elevation  of  it,  as  it  appears  when 
seen  in  the  direction  of  the  arrow.  The  point  T"  then  is  the 
auxiliary  vertical  projection  of  the  lower  front  edge,  NY — T'A'. 
Seen  in  the  direction  of  the  arrow,  the  point,  B,B',  of  the  ray 
AB — A'B',  appears  at  a  distance,  A's,  below  T" — T'A',  and  at 
a  distance,  rB,  to  the  left  of  NY— r" .  Therefore,  make  TV 
equal  to  AY,  and  r"B"  equal  to  rB,  then  T"B"  will  be  the 
auxiliary  vertical  projection  of  a  ray,  or  the  trace  of  a  plane  of 
rays  through  NT— T'A'— T'.  This  plane  cuts  from  the  front 
of  the  cyma  reversa  the  line  u" — uu',  the  upper  shadow  on  the 
cyma  reversa.  A  parallel,  and  tangent,  plane  of  rays,  La", 
gives  the  element  of  shade,  n" — nn',  and  its  shadow,  a" — aa'. 
Another  plane  of  rays,  through  the  edge  below  a" — aa',  would 
give  the  shadow  on  the  lower  part  of  the  abacus,  not  shown. 

2°.  The  shadow  of  the  edge  h'g",  PL  XI.,  Fig.  38,  on  all  the 
cylindrical  parts  below,  is  found  as  in  Prob.  IV. 

The  shadow  of  the  edge  which  is  perpendicular  to  the  paper 
at  A',  on  all  parts  below,  is  bounded  by  the  intersection  of  a 
plane  of  rays  through  this  edge  with  those  lower  surfaces.  This 
intersection  will  appear  in  vertical  projection  as  the  vertical  pro- 
jection of  a  ray  at  h'  (43). 

3°.  The  shadow  of  the  lower  front  edge  of  the  abacus,  RT — 
K'T',  PL  XII.,  Fig.  42,  on  the  echinus  is  found  thus :  Let  N'K' 
— ~Ney  be  a  circle,  cut  from  the  echinus  by  a  horizontal  plane, 
N'K'.  A  ray,  cd — c'o",  through  any  point,  as  c,c',  of  the  edge 
casting  the  shadow,  will  meet  this  plane  in  dd',  which  determines 
dg,  the  shadow  of  RT— R'T'  on  the  parallel  plane,  N'K'  (5). 


96  GENERAL   PROBLEMS. 

This  auxiliary  shadow  meets  the  circle,  Neg — N'K',  at  e,e'  and 
g,g',  two  points  of  the  required  shadow,  which  is  thus  found  by 
(42) — see  also  Prob.  XXX. 

Observing  that  un  is  equal  to  Sf,  the  highest  point,  /',  of  this 
shadow,  may  be  thus  found.  The  plane  of  rays,  through  ET — 
R'T',  intersects  the  meridian  plane,  Dn,  in  aline,  nb,  which  is  the 
projection  of  the  light  on  the  plane  Dn.  The  intersection  of 
this  trace  nb  with  the  echinus  is  the  point/7.  Revolve  bn  to  ba", 
whose  vertical  projection  is  a'b'.  Draw  R'L',  parallel  to  a'b', 
and  revolve  L  to  /',  and  f  is  the  required  highest  point  of  the 
shadow.  By  drawing  a  ray  through  /',  we  could  find  the  point 
whose  shadow  is  f. 

THe  shadow  of  RT — R'T'  is  real,  above  its  intersection  with 
the  curve  of  shade  of  the  echinus,  which  is  found  as  in  Prob. 
XXVII. 

The  remaining  shadows  involve  no  new  operations. 


SHADES  AND  SHADOWS.  97 


SERIES  II. 

SHADOWS  DETERMINED  BY  DIVERGING  RAYS. 

SECTION  I. 
G-eneral    Principles. 

85.  Having  seen,  in  the   preceding  general  problems,  how 
shades  and  shadows  are  found,  when  the  rays  of  light  are  parallel, 
it  now  remains  to  examine  Arts.  (2)  and  (16)  somewhat  in  detail. 
"When  a  body,  B,  is  illuminated  by  a  single  luminous  point,  that 
point  may  be  considered  as  the  vertex  of  a  cone  of  rays,  C,  tan- 
gent to  the  body,  B.     The  line  of  contact  of  C  and  B,  is  the  line 
of  shade  of  B.     The   interior  of  the  cone,  beyond  B,  is  the 
shadow,  in  space,  of  B,  and  that  portion  of  any  secant  surface, 
S,  which  is  within  the  cone,  and  beyond  B,  is  the  shadow  of  B 
on  S.     If,  now,  B  be  illuminated  simultaneously  by  two  points, 
P  and  P',  a  portion,  both  of  B  and  S,  will  receive  light  from 
neither  point.     This  is  the  total  shade  on  B,  or  shadow  on  S, 
respectively.     Another  portion  of  B  and  of  S  will  receive  light 
from  both  points.     This  will  be  their  completely  illumined  por- 
tion.    A  third  portion  of  B,  and  of  S,  will  receive  light  from 
one  or  the  other  of  the  points,  P  and  P',  but  not  from  both. 
This  is  their  penumbra,  or  partial  shade,  or  shadow. 

86.  Let  this  case  be  extended  to  an  assemblage  of  luminous 
points,  forming  a  luminous  body,  L.     In  general,  L  and  B  will 
be  of  different  sizes,  hence  they  may  have  two  common  tangent 
cones  of  rays  inclosing  them ;  one  with  its  vertex  beyond  both 
bodies,  the  other  with  its  vertex  between  the  two  bodies.     The 
area,  or  zone,  on  B,  between  the  curves  of  contact  of  the  two 
cones,  will  be  the  partial  shade  of  B.     The  annular  space  on  S, 
between  the  intersections  of  the  two  cones  with  S,  will  be  the 
penumbra  or  partial  shadow  of  B  on  S — the  part  which  is  reached 
by  only  a  part  of  the  rays  from  L. 

7 


08  GENERAL    PROBLEMS. 

87.  See,  now,  PI.  XIII.,    Fig.  44,  where   AC   represents   a 
luminous  body  ;  BD,  an  opaque  body,  and  PQ  an  opaque  sur- 
face which  receives  the  shadow  of  BD.     The  dark  space,  6c?,  is 
the  total  shadow,  and  is  bounded  by  the  cone  of  rays,  tangent  to 
both  bodies,  and  having  its  vertex  on  the  same  side  of  both. 
The  lighter  space,  ca,  is  the  penumbra,  and  is  bounded  by  the 
tangent  cone  whose  vertex  is  L. 

88.  Eegarding  L  at  first  as  the  original  source  of  light,  and  the 
vertex  of  the  single  cone  Lea,  it  may  then  be  regarded  as  the 
vertex  merely  of  a  complete  cone  of  two  nappes,  in  the  outer 
nappe  of  which  is  inscribed  the  extended  source  of  light,  AC, 
around  which,  and  the  body  BD,  may  be  inscribed  the  second 
cone  which  determines  the  shadow  bd. 

89.  It  should  be  noted,  that  the  volumes  of  rays  will  only  be 
cones,  when  the  bodies,  AC  or  BD,  are  similar,  or  have  similar 
curves  of  contact ;  or  when,  at  least,  they  can  both  be  inscribed 
in  one  given  cone,  so  as  to  have  a  continuous  curve  of  contact 
with  that  cone. 

In  other  cases,  the  volumes  of  rays  will  be  bounded  by  warped 
surfaces,  but  these  cases  are  unimportant,  and  will  not  be  further 
considered.  In  every  case  whatever,  the  surface  of  rays  will  be 
a  ruled  surface,  since  rays  of  light  are  straight  lines. 

90.  For  the  practical  case,  let  L  be  the  sun,  and  B  a  body 
near  the  earth,  S.     Here  the  angular  breadth  of  the  penumbra 
of  shade  upon  a  spherical  body,  B,  would  be,  as  found  by  a 
simple  calculation,  only  T^j-  of  the  radius  of  B. 

See  PL  XIIL,  Fig.  43,  where  O  is  the  sun's  centre,  a,  the 
centre  of  a  small  spherical  body  near  the  earth,  and  V,  and  v, 
the  vertices  of  the  two  tangent  cones  of  rays  already  described. 
Then,  on  account  of  the  relatively  great  distance  of  the  sun, 
AB,  in  the  triangle  ABC,  becomes  sensibly  equal  to  his  diame- 
ter, and  aC  and  ab,  radii  of  the  small  body,  being  perpendicular 
to  BC  and  AC,  respectively,  the  chord  6C  sensibly  reduces  to 
the  segment,  6C,  of  the  side  AC,  and  ABC  and  o&C  are  similar 
triangles,  and  AB  :  BC::Z>C  :  aC  :  or 

885  000  :  95  000  000::iC  :  aC ;  or 
ftC  =  aC         very  nearly. 
Tu7 

This  penumbra  of  shade,  whose  breadth  is  Z>C,  may  therefore 
be  disregarded  on  terrestrial  objects,  and  the  solar  rays  ma}', 
accordingly,  be  considered  parallel. 


SHADES  AND   SHADOWS.  99 

Remark. — The  three  following  problems,  embracing  both 
shades  and  shadows  upon  a  variety  of  surfaces,  plane,  single- 
curved,  and  double-curved,  are  meant  to  be  sufficient  to  initiate 
the  student  in  the  solution  of  problems  in  which  the  source  of 
light  is  supposed  to  be  a  near  luminous  point. 

After  the  explanations  already  made,  separate  statements  of 
"principles"  will  not  precede  the  "constructions"  of  these 
problems. 

SECTION  II. 

!Pro"blems,    involving    divergent   Rays. 

PROBLEM  XXXIII. 

To  find  the  shadow  of  a  semi-cylindrical  abacus,  upon  a  vertical 
plane  through  its  axis. 

PL  XIII.,  Fig.  45,  FcE— F'c'E'  is  the  half  abacus,  with  hori- 
zontal semi-circular  bases,  whose  diameters  are  in  the  vertical 
plane  of  projection. 

L,I/  is  the  source  of  light.  A  line  from  this  point  to  any 
point  of  an  edge,  or  element,  of  shade  of  the  abacus,  is  a  ray  of 
light,  whose  intersection  with  the  vertical  plane  of  projection, 
will  be  a  point  of  the  required  shadow. 

LD  is  the  horizontal  trace  of  a  vertical  tangent  plane  of  rays, 
which  determines  the  element  of  shade  d — d'd" .  Then  from 
d,d"  to  E,E',  is  the  edge  of  shade  of  the  upper  base,  and  from 
djd'  to  F,F',  is  the  edge  of  shade  of  the  lower  base. 

These  edges,  and  the  element  of  shade,  cast  the  line  of  shadow 
which  is  required. 

The  shadow  then  begins  at  F,F'.  Any  point,  as  a,a',  casts  a 
point  of  shadow  A, A',  which  is  found  where  the  ray  La — LV 
pierces  the  vertical  plane  of  projection. 

The  element  of  shade,  d — d'd",  casts  the  shadow  D'D",  and 
the  curve  D"E'  is  the  shadow  of  dE—d"W.  The  shadow  of 
c,c',  the  foremost  point  of  the  lower  edge  of  shade,  is  C',  the 
lowest  point  of  the  shadow  ;  which  is  now  fully  determined. 

The  shadow  of  a  tangent  at  c,c'  will  be  a  tangent  at  C',  parallel 
to  the  ground  line.  Other  obvious  tangents,  useful  in  sketching 
the  curve,  can  readily  be  determined. 


100  DIVERGENT  RAYS. 


PROBLEM  XXXIV. 

To  find  the  curve  of  shade  on  a  sphere,  the  light  proceeding  from  an 

adjacent  point. 

PI.  XIIL,  Fig.  46.  Let  the  centre  of  the  sphere,  Sa'cfc,  be  in 
the  ground  line,  and  let  L,L'  be  the  luminous  point. 

1°.  To  find  the  highest  and  loivest  points  of  shade. — LO  is  the 
horizontal  trace  of  a  vertical  plane  of  rays  containing  these 
points.  After  revolving  this  plane  about  the  vertical  diameter 
of  the  sphere,  till  it  coincides  with  the  vertical  plane  of  projec- 
tion, the  luminous  point  will  appear  at  I/',!/",  and  the  great 
circle  cut  from  the  sphere,  at  Sa'dc.  The  revolved  rays,  I/"A'", 
~L'"l'",  tangent  to  this  circle,  then  determine  h"',h",  and  I'",  the 
revolved  positions  of  the  required  points.  In  the  counter  revo- 
lution, h'",h"  proceeds  in  the  horizontal  arc,  h"h — h'"h'  to  its 
true  position  h,h'.  The  construction  of  1,1',  the  lowest  point, 
will  be  given  presently. 

2°.  To  find  the  foremost  and  hindmost  points. — Go  through  a 
series  of  operations,  similar  to  the  foregoing,  upon  a  plane  of 
rays,  I/O,  perpendicular  to  the  vertical  plane  of  projection, 
beginning  by  revolving  it  into  the  horizontal  plane  of  projection, 
about  the  horizontal  diameter,  whose  vertical  projection  is  0. 
This  will  give  the  points/,/7,  and  e,e'.  The  middle  point,  n,  of 
f'e'j  is  the  vertical  projection  of  the  centre  of  the  circle  of  shade, 
then  I',  the  vertical  projection  of  the  lowest  point,  is  at  the  inter- 
section of  Z"T,  the  vertical  projection  of  an  arc  of  counter  revo- 
lution, with  the  diameter,  h'n'l'.  Then  I'  is  horizontally  projected 
in  LO,  at  I. 

3°.  The  points  on  tfie  circles  which  are  in  the  planes  of  projection, 
are  found  by  drawing  tangent  planes  of  rays,  perpendicular  to 
the  planes  of  projection.  As  the  centre  of  the  sphere  is  in  the 
ground  line,  one  circle  of  the  diagram  represents  both  of  its  pro- 
jections. Then  IS  a'  and  L'i'  are  the  vertical  traces  of  tangent 
planes  of  rays  perpendicular  to  the  vertical  plane.  They  give 
a',a  and  b',b,  as  the  points  of  shade,  on  the  circle,  SZ> — Sa'c?c? 
which  is  in  the  vertical  plane  of  projection. 

Likewise,  Lc  and  ~Ld,  the  horizontal  traces  of  vertical  tangent 
planes  of  rays,  give  the  points  c,c'  and  d,df  on  the  circle  Sa'dc — 
Si,  which  is  in  the  horizontal  plane  of  projection. 


SHADES  AND   SHADOWS.  101 

The  construction  of  other  points,  as,  for  instance,  those  on  the 
great  circle,  perpendicular  to  the  ground  line,  is  left  for  the 
student. 

Having  now  eight  points  of  shade,  the  curve  of  shade  may 
be  sketched.  To  avoid  confusion,  only  the  vertical  projection 
of  the  shade  is  shown,  the  horizontal  projection  of  the  points 
of  the  curve  of  shade  being  left  unconnected.  Also,  the  whole 
shade  on  the  front  hemisphere,  is  represented  as  visible,  for  the 
sake  of  clearness,  though  only  a  quadrant  of  the  sphere  is  in  the 
first  angle. 

PROBLEM  XXXV. 

Having  a  niche,  whose  base  is  produced,  forming  a  full  circle;  and 
a  right  cone,  the  centre  of  whose  circular  base  coincides  with  the 
centre  of  the  base  of  the  niche,  it  is  required  to  find  the  shades  and 
shadows  of  this  system,  when  illumined  by  an  adjacent  point. 

1°.  The  shadows  on  the  base  and  cylindrical  part. 

PI.  XIII.,  Fig.  47.  The  given  magnitudes  being  familiar 
ones,  their  projections,  in  the  position  described,  may  be  under- 
stood from  the  figure.  L,L'  is  the  luminous  point. 

La  is  the  horizontal  trace  of  a  vertical  plane  of  rays,  through, 
the  edge  A — A' A"  of  the  niche,  and  limits  the  shadow,  Aa,  of 
the  niche  upon  its  base.  From  a,a",  the  shadow  of  the  same 
edge  is  the  element,  a — a"a',  of  the  cylindrical  part  of  the  niche, 
limited  by  the  ray  La — L'a". 

The  shadow,  e,e',  of  a  point,  F,F',  of  the  front  circle  of  the 
spherical  part,  upon  the  cylindrical  part,  is  seen  to  be  found  as 
in  Prob.  XVIIL,  the  ray  being  drawn  through  L,L'. 

2°.  The  shadow  of  the  spherical  part  on  itself,  is  found  essentially 
as  in  Prob.  XXIX.,  only  each  point  of  this  shadow  requires  a 
separate  auxiliary  plane,  whose  vertical  trace  will  pass  through 
L'.  Thus,  L'N'  is  the  trace,  on  the  front  face  of  the  niche,  of  an 
auxiliary  plane  perpendicular  to  the  vertical  plane  of  projection. 
The  ray,  in  this  plane,  meets  the  semicircle,  cut  by  it  from  the 
niche,  in  a  point  of  shadow.  Wl"N',  described  on  M'N'  as  a 
diameter,  is  the  semicircle  just  named,  after  revolution  about 
M'N'  into  the  front  face  of  the  niche.  The  point  L,L'  is  at  a 
distance,  LJ,  from  the  front  of  the  niche,  hence,  making  L'L" 
equal  to  LJ,  and  perpendicular  to  L'N',  it  results  that  L"MT' 
is  the  revolved  position  of  a  ray,  giving  I"  for  the  revolved 


102  DIVERGENT  RAYS. 

position  of  a  point  of  shadow.  By  counter-revolution  about 
L'N',  the  point  I"  returns  to  ?',  its  true  position  as  the  shadow 
ofM'. 

3°.  The  method  of  Prob.  XXIX.  (2d  solution)  is  here  shown, 
also.  Thus,  let  DE  be  the  horizontal  trace  of  a  vertical  plane, 
parallel  to  the  front  of  the  niche.  It  cuts  from  the  spherical 
part,  the  semicircle  DE— D/'E'.  The  ray,  LV— I/O',  through 
the  centre  of  the  front  semicircle,  AYB,  pierces  this  plane  at  0,0'. 
The  horizontal  line,  o'a'",  of  this  plane,  limited,  at  a'",  by  the 
ray  L' A'V",  is  the  shadow  of  the  radius  O'A"  on  the  plane  DE. 
Hence  a'"f,  with  centre  o',  and  radius,  o'a'",  is  an  arc  of  the 
shadow  of  the  front  circle,  on  DE,  and  /',  its  intersection  with 
the  semicircle  D/'E',  cut  from  the  niche  by  the  plane  DE,  is  a 
point  of  shadow  on  the  spherical  part. 

It  will  now  be  easy  to  find  the  point  n' ;  the  point  of  which 
f  is  the  shadow ;  and  the  horizontal  projection  of  the  curve 
T7V,  remembering  that  it  is  a  plane  curve  (88). 

4°.  To  find  the  elements  of  shade  of  the  cone,  and  its  shadows. 
Draw  the  ray  LV — LV''  and  find  U,  where  it  pierces  the  plarut) 
of  the  base  of  the  niche.  U£  and  ~Uq  are  the  horizontal  traces 
of  planes  of  rays,  tangent  to  the  cone,  and  tV — t'V,  and  qV — qf 
V,  are  its  elements  of  shade. 

The  traces  Utf  and  U^  bound  the  shadow  on  the  base  of  the 
niche.  At  ni,m'  and  p,p'  the  shadow  on  the  cylindrical  part  of 
the  niche  begins.  The  ray  LV — L'V  pierces  this  part  at  v,v',  the 
shadow  of  the  vertex  on  the  interior  of  the  niche.  The  shadows 
of  the  elements  of  shade  being  curves,  other  points  besides  m,m' 
and  p,p'  muse  be  found.  To  avoid  the  acute  intersections  of  pro- 
jecting lines  with  q'V  and  t'V,  find,  by  division,  their  middle 
points,  giving  <7,#' and  h,h'.  Kays,  L^ — L'#' and  Ui — L'/i',  through 
these  points,  pierce  the  cylinder  of  the  niche,  produced,  at  s,s' 
and  r,r',  respectively.  Through  these,  and  the  previously  found 
points,  the  indefinite  shadows,  v'm's'  and  v'p'r',  of  the  elements  of 
shade,  can  be  drawn.  The  points,  as  u,  where  the  horizontal  pro- 
jections of  the  rays  just  drawn  meet  the  horizontal  traces  of  the 
planes  of  shade,  are  the  points  in  which  those  rays  pierce  the  hori- 
zontal plane,  and  should  therefore  appear  in  perpendiculars  to  the 
ground  line  through  u',  etc.,  which  are  their  vertical  projections. 

The  niche  conceals  the  horizontal  projections  of  the  shades 
and  shadows  except  the  portion  of  shade  V/K  on  the  cone,  and 
tlcK,  a  portion  of  the  shadow  of  the  cone,  on  the  base  of  the  niche. 


SHADES  AND   SHADOWS.  103 


PART  II. 
SHADES  AND  SHADOWS  IN  ISOMETRICAL  PROJECTION. 

> 

SECTION  I. 
General    Principles. 

91.  Isometrical  projections  of  shadows  may  be  found  by  two 
quite  different  methods. 

First — They  may  be  found  directly  on  the  isometrical  projec- 
tion of  the  object  which  receives  them.  Second. — They  may  be 
first  constructed  in  ordinary  orthographic  projection,  and  then, 
from  such  projections,  their  isometrical  projections  may  be  found. 
Examples  of  both  of  these  modes  of  procedure  will  be  given 
among  the  following  problems. 

92.  Since  the  essential  feature  of  isometrical  drawing  is,  that 
it  shows  three  dimensions,  at  right  angles  to  each  other,  in  their 
real  size,  it  follows,  that  this  kind  of  drawing  is  chiefly  useful  in 
its  applications  to  rectangular  bodies.     This  fact,  also,  will  be 
illustrated  in  the  following  constructions. 

93.  In  isometrical  projection,  only  one  principal  plane  of  pro- 
jection is  used  ;  but,  according  to  first  principles,  two  projections 
of  a  line,  i.  e ,  its  position  with  respect  to  at  least  two  planes,  are 
necessary  to  determine  its  position  in  space  ;  hence,  in  isometrical 
projection,  the  rays  of  light  are  more  conveniently  determined  by 
referring  them  to  isometrical  surfaces,  as  those  of  the  body  illu- 
minated, than  to  planes  of  projection.     Hence,  again,  isometrical 
drawing  is  most  useful  in  connection  with  rectangular  objects ; 
since,  in  representing  other  objects,  auxiliary  projections  of  rays 
must  be  employed,  as  will  soon  be  seen. 

In  the  following  problems  only  parallel  rays  are  employed. 
In  PI.  XIV.,  Fig.  48,  the  light  is  supposed  to  enter  the  cube  at 
the  upper  left  hand  corner,  E,  and  to  leave  it  at  the  lower  right 
hand  corner,  J,  and  this  is  the  conventional  direction  of  the  light, 
generally,  in  isometrical  drawing. 


101  ISOMETRICAL   SHADOWS. 

The  ray,  being  thus  a  diagonal  of  the  cube,  makes  an  angle  of 
35° — 16'  with  each  of  its  faces.  Its  projections,  as  EK  and  ET, 
on  those  faces,  are  diagonals  of  the  faces,  and  therefore  make 
angles  of  45°  with  the  edges  of  the  cube.  In  the  absence  of 
these  planes  and  edges,  as  when  a  curved  surface  is  the  subject 
of  a  problem,  it  may  be  convenient  to  know  the  angle  made  by 
the  ray  with  the  plane  of  projection.  This  we  next  turn  to  find. 


PROBLEM  XXXVI. 

To  find  the  angle  made  by  the  isometrical  ray  of  light  with,  the 
isometrical  plane  of  projection. 

The  point,  C,  Fig.  48,  is  the  projection  of  the'  diagonal  from 
the  foremost  to  the  hindmost  points  of  the  cube,  hence  the  plane 
of  projection  is  perpendicular  to  this  diagonal.  Then,  in  Fig.  49, 
let  ACBD  be  the  plan  of  a  cube,  and  D'C'E'G'  its  elevation ; 
the  cube  being  in  the  fourth  angle,  with  its  vertical  faces  making 
equal  angles  with  the  vertical  plane  of  projection,  parallel  to  the 
paper.  Then  PQ,  perpendicular  to  the  diagonal  D'F',  is  the 
trace,  on  the  plane  of  the  paper,  of  the  plane  of  isometrical  pro- 
jection, and  AB  and  B'E'  are  the  two  projections  of  the  ray  seen 
at  EJ,  in  Fig.  48.  Observe  that  EK,  El,  and  KI,  Fig.  48,  are 
parallel  to  the  isometrical  plane  of  projection.  Then  B'G',  Fig. 
49,  is  the  trace  of  a  plane,  parallel  to  the  isometrical  plane  PQ, 
and  the  point  A,B'  is  its  own  projection  on  this  plane,  and  the 
point  B,E'  is  projected  upon  it  by  the  perpendicular  E'<7,  which 
is  seen  in  its  true  size.  But  the  diagonals  of  the  cube  being 
equal,  D'F'=  the  line  AB— B'E',  also  E'#  =  i  of  D'F',  and  hence 
of  the  ray  AB — B'E',  and  if  then  the  ray,  considered  as  radius, 
be  called  1,  E'<7  =  £  will  represent  the  sine  of  the  angle  made  by 
the  ray  with  the  parallel  plane  B'G'  and  hence  with  the  isome- 
trical plane  PQ.  But  1  is  the  Nat.  sine  of  19° — 28',  very  nearly, 
which  is  therefore  the  desired  value  of  the  angle  made  by  a  ray 
of  light  with  the  plane  of  isometrical  projection. 


-.n- 
SHADES   AXD  SHADOWS.  lOo 

SECTION  II. 
Shad.es    and.    Shadows    on    Isornetrical    Planes. 

94.  Shadows  on  isometrics!  planes  will  be  the  intersections  of 
rays,  through  points  casting  shadows,  with  their  projections  on 
such  planes,  or  with  the  traces,  on  the  same  planes,  of  any  planes 
of  rays  containing  those  points. 

PROBLEM  XXXVII. 

Having  given  a  cube,  with  thin  plates  projecting  vertically  and  for- 
ward, in  Hie  plane  of  its  left-hand  back  face,  to  find  the  shadows 
of  the  edges  of  these  plates  upon  the  cube  and  its  base. 

PI.  XIV.,  Fig.  48.  To  find  the  shadow  of  AB  upon  the  top  of 
the  cube,  to  which  it  is  parallel. — Acr  and  B6  represent  the  rays 
themselves,  through  A  and  B  ;  hence,  as  AE  and  Be  are  project- 
ing linen,  perpendicular  to  the  top  of  the  cube,  E#  and  c&,  on, 
and  parallel  to  EK,  are  the  projections  of  these  rays,  upon  the 
face  ECKL.  These  rays  meet  their  projections  at  a  and  5,  the 
shadows  of  A  and  B,  hence  ab  is  the  shadow  of  AB.  EaJc  is 
now  evidently  the  shadow  of  the  rectangle  ABcE. 

To  find  the  shadow  o/"EDFG. — Here  DE,  being  perpendicular  to 
the  face  ECIGr,  its  shadow  on  that  face  is  in  the  trace,  El,  of  a 
plane  of  rays,  through  DE,  upon  that  face,  and  is  limited  at  c?,  by 
the  ray  T)d.  From  d,  dh,  parallel  to  DF,  is  the  shadow  of  the 
portion,  DH,  of  the  edge  DF.  Then  JiF  is  the  shadow  of  HF 
on  the  plane  of  the  lower  base  of  the  cube. 

We  see  that  PI.  XIV.,  Fig.  50,  shows  the  case  in  which  the 
line,  AB,  casting  the  shadow,  coincides,  in  projection,  with  a  ray. 
Either  point,  as  6,  of  the  shadow,  may  be  found,  either  by  pass- 
ing a  plane  of  rays  through  B,  and  perpendicular  to  the  top  sur- 
face FHK,  or  to  the  face  FKI.  The  line  BD,  perpendicular, 
and  the  trace  Db  parallel  to  FI,  determine  a  plane  in  the  former 
position,  and  6,  the  intersection  of  the  trace  D6,  and  ray  BZ>,  is 
the  shadow  of  B.  BE,  parallel  to  KH,  and  the  trace  E6  in  the 
direction  of  El,  Fig.  48,  determine  a  plane  in  the  latter  position, 
above  named ;  and  b  is  seen  to  be,  as  before,  the  shadow  of  B 
on  FKI. 


106  ISOMETRICAL   SHADOWS. 

Observe  that  Eb,  and  El,  Fig.  48,  make  angles  of  60°  with  a 
horizontal  line. 


SECTION  III. 
Sb.ad.es    and.    Shadows    on.    N"on-Isometrical    3?lanes.' 

PROBLEM  XXXVIII. 

To  find  the  shadow  of  a  hexagonal  cupola  on  a  coupled  roof,  one 
face  of  the  cupola  making  equal  angles  with  two  adjacent  walls  of 
the  house. 

PL  XIV.,  Fig.  51.  To  locate  the  walls  and  roof  Let  A'"A"Y 
and  YA'V  be  portions  of  the  adjacent  walls,  which  are  at  right 
angles  to  each  other. 

Suppose  the  inclination  of  the  roof  to  the  horizontal  plane, 
A'"A'W,  of  the  eaves,  to  be  30°.  Find  0,  the  middle  point  of 
A'" A".  Then  make  OA',  horizontal,  and  equal  to  OA'",  and 
make  OA/B/=30°,  then  B',  the  intersection  of  the  vertical  OB', 
with  A'B',  is  the  extremity  of  the  summit  of  the  roof;  through 
it  B'A"',  B'A",  and  B'M,  may  be  drawn,  and  through  M,  the 
ends  of  the  roof,  parallel  to  A"'B'  and  A"B'. 

1°.  To  locate  the  cupola. — Let  n  be  the  point  at  which  its  axis, 
riN,  pierces  the  centre  line,  On,  of  the  plane  of  the  eaves,  and 
let  nv  be  the  half  width  of  the  square  vgde,  in  which  the  circum- 
scribing circle,  acbe,  of  the  base  of  the  cupola  is  inscribed.  The 
semicircle,  abb,  represents  the  semicircle,  ac5,  after  revolution 
about  the  axis  ai,  till  it  is  parallel  to  the  plane  of  the  paper,  i.  e., 
to  the  isometrical  plane  of  projection.  Hence  trisect  akb,  and 
at  k  and  h  draw  Ice  and  Ac?,  the  projections  of  the  arcs  of  coun- 
ter-revolution described  by  k  and  h;  and  ac,  cd,  and  db,  will  be 
three  sides  of  the  base  of  the  cupola.  The  remaining  sides  are 
parallel  to  these  three.  At  the  corners  a,  c,  etc.,  of  the  base, 
draw  the  equal  vertical  edges  aC,  cl,  etc.,  and  join  their  upper 
extremities,  which  will  complete  the  cupola. 

2°.  To  construct  the  intersection  of  the  cupola  with  the  roof. — The 
plane  of  the  face  led  gives  the  trace,  urn,  on  the  plane  of  the 
eaves,  and  the  trace  uU,  on  the  vertical  plane  through  the  ridge 
B'M.  Then  Urn  is  its  trace  on  the  front  roof,  and  CD,  the  defi- 
nite intersection  of  the  face  led  with  the  front  roof.  Similar 


SHADES  AND   SHADOWS.  107 

planes,  through  Hi.  and  Gcr,  give  the  traces  Nn,  on  the  front 
roof,  and  NP,  on  the  back  roof,  and  give  B  and  A,  as  the  inter- 
sections of  these  edges  with  those  roofs. 

The  vertical  plane  through  the  ridge,  cuts  the  edges,  ac  and 
be,  of  the  base,  at  s  and  t.  Vertical  lines,  sS,  and  iT,  from  these 
points,  meet  the  ridge  at  S  and  T,  where  it  meets  the  walls  of 
the  cupola.  From  S  and  T,  draw  S  A  and  SO ;  TB  and  TE, 
which,  with  AF  and  BD,  complete  the  required  intersection. 

3°.  To  find  the  shades  and  shadows. — The  auxiliary  planes,  just 
used,  being  planes  of  rays,  determine  the  faces  whose  upper 
edges  are  IJ,  JH,  HK,  and  KL,  as  in  the  shade.  Kays,  as  Ir 
and  Hp,  through  the  extremities  of  these  edges,  meet  the  traces, 
O  and  Bp,  of  the  planes  of  rays  on  the  roof,  produced,  at  r,  p, 
etc.  Then  Drq'poQE  is  the  required  shadow,  of  which  only 
the  part  on  the  actual  roof  is  real. 

The  student  may  reconstruct  the  figure,  with  the  face  ID 
parallel  to  the  wall  YA"?i. 


SECTION  IV. 

Shades    and.    Shadows    on.    Single    Curved    Surfaces. 

PROBLEM  XXXIX. 

To  find  the  elements  of  shade  on  an  inverted  hollow  right  cone,  the 
shadow  on  its  interior,  and  the  shadow  of  one  of  its  elements  of 
shade  on  an  oblique  plane. 

1°.  To  find  the  elements  of  shade.  PI.  XIY.,  Fig.  52.— Let  the 
isometrical  ellipse,  SQTJST,  be  the  upper  base,  0V  the  axis,  and 
the  tangents,  VQ  and  VN,  the  extreme  elements  of  the  cone. 
YOR  is  a  plane  of  rays,  and  its  trace,  OR,  on  the  base,  meets 
the  ray  YR,  parallel  to  E  J,  Fig.  48,  at  R,  the  intersection  of  a 
ray  through  the  vertex,  with  the  plane  of  the  base.  Then  RS 
and  RT  are  the  traces,  on  the  plane  of  the  base,  of  tangent  planes 
of  rays,  which  determine  TV  and  SV  as  the  elements  of  shade. 

2°.  To  find  the  shadow  on  the  interior. — R6  is  the  trace  of  a 
secant  plane  of  rays,  which  cuts  from  the  base  the  point  a,  cast- 
ing a  shadow,  and  from  the  cone,  the  element,  6Y,  receiving  the 
shadow.  Then  the  ray,  a  A,  determines  A,  the  shadow  of  a, 


108  ISOMETRICAL  SHADOWS. 

In  the  same  way,  other  points,  C,  etc.,  may  be  found.  The 
shadow  ends  at  S  and  T.  The  shadow  of  e  is  the  lowest  point 
of  shadow,  e  being  the  point  furthest  from  the  element  which 
receives  its  shadow.  The  shadow  of  e  is  E,  on  the  element /V 
(undistinguishable  in  the  figure  from  NV).  The  shadow  on 
the  element  NV,  is  found  by  drawing  a  trace  EN. 

3°.  To  find  the  shadow  of  the  front  element  of  shade  on  an  oblique 
plane,  BLFL — DKFI  is  in  an  assumed  horizontal  plane  through 
FI ;  and  as  its  position,  relative  to  the  cone,  is  undetermined  by 
the  projection,  it  is,  when  produced,  assumed  as  cutting  the  axis 
0V  at  m.  Then  make  Tt  equal  to  Om,  and  t  is  the  projection 
of  T  on  the  plane  DKFI.  Hence  tu  is  the  trace,  on  this  plane, 
of  a  vertical  plane  of  rays  through  T ;  wU  is  its  trace  on  the 
vertical  surface  BDF,  and  Uy,  on  the  oblique  plane.  The  ray, 
Ty,  therefore  gives  y,  as  the  shadow  of  T. 

Another  point  of  the  shadow  of  TV,  is  where  it  pierces  the 
oblique  plane.  Now  tm,  parallel  by  construction  to  TO,  is  the 
trace  of  a  meridian  plane,  OTtm,  on  the  horizontal  plane  DFI. 
This  meridian  plane  being  vertical,  it  intersects  the  vertical  sur- 
faces, BDF  and  BDK,  in  the  vertical  lines  AH  and  gGc,  giving 
GH  for  its  trace  on  the  oblique  plane.  But  this  meridian  plane 
contains,  by  construction,  the  element  of  shade,  TV.  Hence  r, 
where  TV  meets  the  trace  GHr,  is  the  intersection  of  TV  with 
the  oblique  plane,  produced,  and  is  therefore  a  point  of  its  sha- 
dow on  that  plane.  Hence  ry  is  the  required  shadow  of  TV. 

The  shadow  of  SV  may  be  similarly  found. 

The  distance  equal  to  Om,  might  first  be  assumed  as  at  pP, 
then  Pin,  parallel  to  pO,  will  locate  m,  and  t  will  be  the  intersec- 
tion of  Tt  with  P*,  parallel  to^T. 


SECTION  V. 

Sh.ad.es    and.    Shadows    on    Double-Curved    Surfaces. 

PKOBLEM  XL. 
To  find  the  curve  of  shade  on  a  sphere.. 

The  projection  of  the  sphere  on  the  isometrical  plane  of  pro- 
jection is  the  circle  CBDA,  PI.  XIV.,  Fig  53.  Make  the  angle 
KF'E'  equal  to  PF'C',  Fig.  49,  then  KF'  will  be  the  intersec- 


SHADES   AND   SHADOWS.  109 

tion  of  the  plane  of  the  paper,  which  is  the  isometrical  plane  of 
projection,  with  an  auxiliary  plane,  taken  as  a  vertical  plane  of 
projection,  and  on  which  E'F'  is  the  trace  of  a  plane  of  rays. 
Project  back  O  at  0',  and  make  O'E^O'F',  and  each  equal  to 
OA.  Then  E'F'  is  the  auxiliary  projection  of  that  great  circle 
which  is  contained  in  a  plane  of  rays.  Project  E'  and  F'  at  E 
and  F  ;  then  the  ellipse  on  AB  and  EF  as  axes,  will  be  the  iso- 
metrical  projection  of  the  great  circle  just  named.  Planes  of 
rays,  parallel  to  E'F',  will  cut  parallel  small  circles  from  the 
sphere,  whose  diameters  will  be  chords  of  the  horizontal  great 
circle,  whose  isometrical  projection  is  AHBG. 

As  these  parallel  circles  will  be  projected  in  similar  ellipses, 
assume  gh,  kn,  etc.,  as  their  diameters,  and  draw  hs  and  nr  parallel 
to  BE,  which  will  give  the  conjugate  axes,  es  and  fr,  of  these 
ellipses.  Constructing  these  ellipses,  and  drawing  rays  tangent 
to  them,  gives  points  of  shade,  d,  c,  &,  etc. 

The  tangent  at  <z,  parallel  to  the  rays,  is  the  trace  of  a  plane 
of  rays  perpendicular  to  the  plane  of  isometrical  projection,  and 
gives  a  as  the  point  of  shade  on  the  great  circle,  parallel  to  the 
paper,  which  forms  the  isometrical  projection  of  the  sphere. 

H  and  Gr  are  points  of  shade,  and  a  line  aOp  is  the  trace  of  the 
plane  of  shade  on  a  meridian  plane  parallel  to  the  paper,  giving 
p  for  another  point  of  shade.  Other  points  can  be  found  as  first 
described. 

JZemark. — For  an  approximate  construction,  circles  with  radii, 
eh,  etc.,  may  be  employed  instead  of  ellipses,  in  ordinary  con- 
structions. 


PROBLEM  XLI. 
To  construct  the  curve  of  shade  on  a  torus. 

PI.  XIY.,  Fig.  54.  KT  is  the  intersection  of  the  isometrical 
plane,  or  plane  of  the  paper,  with  the  auxiliary  vertical  plane. 
FI/  is  the  ground  line  of  the  latter  plane,  on  that  auxiliary  plane 
which  is  used  as  a  horizontal  plane  of  projection,  and  is  found 
by  making  the  angle  I/FT=E'F'Q,  Fig.  49. 

~Lda"  and  a'e'h"u'  are  the  auxiliary  projections  of  the  torus. 
The  auxiliary  projections  of  the  light,  in  the  position  which  it  is 
desired  to  have  on  the  isometrical  projection,  are  0"L— K'L'. 


110  ISOMETRICAL  SHADOWS. 

We  now,  according  to  (91  Second),  construct  the  auxiliary  pro- 
jections of  the  curve  of  shade,  and  then  the  isometrical  projection 
of  the  torus  and  its  shade. 

1°.  To  find  four  points  of  the  curve  of  shade. — From  Fig.  48,  it 
appears  that  the  light  makes  an  angle  of  35° — 16'  with  the  plane 
of  the  base  of  the  cube,  i.  e.,  with  the  horizontal  plane  of  projec- 
tion. Then,  knowing  that  the  horizontal  plane  (so  called  in 
Fig.  54)  and  seen  edgewise  before  revolution  at  FL/,  makes  this 
same  angle  with  the  horizon,  it  follows  that  O"L" — R'L'"  is  a 
ray,  after  being  revolved  about  the  (relatively)  vertical  axis 
0" — I/CX,  till  parallel  to  the  auxiliary  vertical  plane  ;  and  R'L'" 
is  found  horizontal.  Then  draw  rays,  tangent  at  e'  and  u',  and 
parallel  at  R'L"',  and  e'  and  u'  will  be  the  revolved  positions  of 
the  highest  and  lowest  points  of  the  curve  of  shade,  whose  true 
positions  are  h'  and  R'. 

The  points  of  shade  on  the  greatest  horizontal  section  of  the 
torus,  are  a", a!  and  b",b'. 

2°.  To  find  intermediate  points. — Let  the  points  in  the  meridian 
plane,  cd,  be  found.  According  to  (80)  O"l  and  RT  represent 
the  ray  0"L—  RT/  after  being  projected  on  this  plane,  cd. 
Then  0"l"  and  RT"  are  the  revolved  positions,  in  the  parallel 
meridian  plane,  a"b",  of  the  projected  ray.  Next,  draw  from  r, 
the  centre  of  the  semicircular  part  of  the  meridian  curve,  rd"f, 
perpendicular  to  RT",  then  d'",d"  is  the  revolved  position,  and 
d,d'  the  primitive  position  of  a  point  of  shade  ;  O"— LVi'  being 
the  axis  of  revolution.  For  variety  of  construction,  revolve  Yc 
to  the  position  Yc",  parallel  to  the  vertical  plane  of  projection. 
This  semicircle  will  then  be  vertically  projected,  with  the  radius 
p'm' ;  and  by  drawing  p'c'",  perpendicular  to  RT",  we  find  the 
point  of  tangency,  c'",c",  °f  a  revolved  ray,  parallel  to  RT",  and 
from  this,  the  same  point  in  its  true  position,  c,c'. 

3°.  To  construct  the  isometrical  projection  of  the  fonts,  and 
of  its  shade.  Assume  eu  for  the  trace  of  the  meridian  plane, 
a"b",  upon  the  isometrical  plane. 

The  highest  points,  e  and  u,  are  projected  from  e'  and  u'  upon 
this  trace.  The  right  and  left  hand  points,  E  and  G,  are  pro- 
jected from  0',  and  EG-  is  the  projection  of  the  diameter  E"G". 
To  find  intermediate  points,  as  those  in  the  plane  cd,  project 
the  direction  of  vision,  sO" — s'O',  upon  this  plane  as  at  SO" — 
S'O'.  Then  revolve  this  projected  ray,  together  with  the  meridian 
plane,  about  a  vertical  axis  at  0"  till  parallel  to  the  vertical 


SHADES  AND  SHADOWS.  Ill 

plane  of  projection,  and  then  by  (80)  the  points  of  contact  of 
projecting  lines,  parallel  to  s'"O',  will  be  points  of  apparent 
contour  in  the  isometrical  projection.  These  points  are  best 
found  by  drawing  radii,  as  rn'",  perpendicular  to  s'"O',  which 
gives  n'"n" ,  and  after  counter-revolution,  n,n'.  There  being 
evidently  four  such  points,  make  bN=ns,  and  then  making 
Oa= Ob,  make  al,  aB,  and  b A,  each  equal  to  bN.  The  oval  figure 
passing  through  the  points  now  found,  will  be  the  isometrical 
projection  of  the  torus. 

For  the  curve  of  shade ;  h'  is  projected  at  H,  tR  being  equal 
to  h'h".  Oq=0t,  and  then,  qR=tK.  The  points  c'  and  d'  are 
then  projected  at  C  and  D,  at  distances  from  eu  equal  to  the 
distance  of  d  from  a"~b".  Finally,  a'  and  b'  are  projected  in  the 
meridian  curve,  a"b"j  whose  isometrical  projection  is  eu,  at  a  ana 
b.  Through  the  six  points  now  found,  with  f  and  &,  the  points 
of  tangency  of  tangent  planes  of  rays  perpendicular  to  the  isome- 
trical plane,  the  curve  of  shade  can  be"1  drawn. 

H,  the  highest  point,  determines  the  visible  portion  of  the 
curve  of  shade  to  be/*HD&. 

Remarks. — a.  After  all  the  labor  of  making  the  foregoing 
construction,  it  is  now  more  fully  evident,  according  to  (92),  that 
while  this,  and  the  three  preceding  constructions,  afford  good 
examples  for  study,  they,  and  the  present  problem  particularly, 
show  that  isometrical  projection  has  little  or  no  advantage  in 
respect  to  clearness  of  representation,  except  as  applied  to  plane- 
sided  bodies  having  solid  right  angles,  whose  sides  are  equally 
inclined  to  the  plane  of  projection. 

b.  On  account  of  the  superior  pictorial  character  of  isometrical 
projections,  and  still  more,  of  the  oblique  projections  explained 
in  a  previous  volume  (Elementary  Projection  Drawing)  under 
the  name  of  "  Cabinet  Projections,"  it  is  of  less  importance  to 
represent  shades  and  shadows  upon  them.  The  cabinet  projec- 
tions of  shades  and  shadows  can  readily  be  made  from  their  com- 
mon projections. 


112  BRILLIANT  POINTS. 


2. 

THE   FINISHED  EXECUTION  OF  SHADES  AND 
SHADOWS. 

CHAPTER  I. 

THEORY  AND  CONSTRUCTION  OF  BRILLIANT  POINTS ; 
AND  OF  GRADATIONS  OF  SHADE. 

SECTION  I. 
Preliminary    General    Principles. 

§  1°. — Geometrical  Conditions  for  the  adequate  graphical  repre 
sentation  of  Form. 

95.  The  obviously  essential  geometrical  feature  of  a  surface  is 
its  continuity.     But  the  bounding  surface  of  a  volume  is  repre- 
sented, geometrically,  as  seen  from  a  given  point,  by  its  apparent 
contour ;  which .  is  only  that  line  of  the  surface  which  is  its 
visible  boundary,  as  seen  from  that  single  point. 

Geometrically,  therefore,  an  infinite  number  of  consecutive 
contours,  seen  from  as  many  points  of  view,  on  each  of  the  two 
opposite  sides  of  a  body,  would  be  necessary  to  completely 
represent  its  continuity  and  conformation. 

But  the  method  of  projections  usually  gives  but  two  apparent 
contours ;  viz.,  those  which  constitute  the  two  projections  of  a 
body.  Furthermore,  it  would  be  graphically  impossible  to 
represent  consecutive  apparent  contours ;  hence,  purely  geo- 
metrical diagrams,  though  enabling  us  to  represent  any  required 
point  of  a  surface,  can  never,  of  themselves,  adequately  repre- 
sent the  continuity  of  the  inclosing  surface  of  a  body. 

96.  It  is  therefore  our  previous  and  definite  conception  of 
the  surfaces  to  be  represented,  which  enables  the  projections  of 


SHADES  AND  SHADOWS.  113 

surfaces  practically  to  express  the  forms  of  those  surfaces  intelligi- 
bly. This  may  be  illustrated  by  the  difficulty  usually  experi- 
enced at  first,  in  comprehending  projections  of  new  surfaces,  as 
warped  surfaces,  before  acquiring,  from  models  or  other  sources, 
some  idea  of  their  form.  On  the  other  hand,  the  circular  and 
rectangular  projections  of  the  familiar  cylinder  of  revolution, 
perfectly  represent  it  to  the  mind,  because  we  so  well  know  that 
all  its  right  sections  are  equal  circles,  and  all  its  meridian  sections 
are  equal  rectangles. 

97.  By  calling  in  physical  considerations  to  the  aid  of  geo- 
metrical ones,  and  confining  ourselves  to  the  sense  of  sight  as  a 
means  of  judging  of  the  configuration  of  bodies,  the  action  of 
light  upon  their  surfaces  would  be  observed  at  once.     This  leads 
to  the  following  considerations  : 

§  2°. — Of  the  physical  conditions  for  the  visibility  of  Bodies ', 
and  an  adequate  representation  of  their  Forms. 

98.  The  most  immediately  obvious  result  of  the  exposure  of 
a  body  to  the  light,  is  the  existence,  upon  the  body,  of  a  line  of 
shade  separating  the  illuminated  from  the  shaded  portion  of  the 
body;  and  the  production  of  a  shadow  upon  any  adjacent  sur- 
face from  which  light  is  excluded  by  toe  given  body.     The  con- 
struction of  the  lines  of  shade  and  of  shadow  have,  accordingly, 
claimed  attention  in  all  the  previous  problems. 

But  from  (95)  the  curve  of  shade,  alone,  is  insufficient,  together 
with  the  apparent  contour,  to  determine,  unequivocally,  the  form 
of  a  body.  This  becomes  further  apparent,  as  follows:  Having 
a  given  cylinder  of  rays,  any  curve  traced  upon  its  surface,  may 
be  the  curve  of  contact  of  an  infinite  number  of  different  bodies, 
all  of  which  shall  have  the  same  projecting  cylinder,  and  all  of 
which  will  have  this  one  assumed  curve  for  their  curve  of  shade. 
All  these  bodies,  being  inscribed  tangentically  in  the  same  cylin- 
der of  rays,  will,  moreover,  cast  identical  shadows  on  any  other 
given  surface. 

99.  See,  at  this  point,  PI.  XV.,  Fig.  55.     NBEC  is  any  opaque 
body.     ABKD  is  a  tangent  cylinder  of  luminous  rays,  giving 
the  curve  of  shade  BHDG.     Next,  IFCJ  represents  a  tangent 
projecting  cylinder,  or  cylinder  of  visual  rays  reflected  from  the 
object,  and  gives  the  apparent  contour  FGrCH.     Hence  the  por- 
tion, E — FGCH,  of  the  body,  bounded  by  that  contour,  and  the 

8 


BRILLIANT   POINTS. 

portion,  GDH,  of  its  curve  of  shade,  are  visible.  But  besides 
the  curve  and  that  contour,  there  is  only  our  imagination  to  sug- 
gest the  real  configuration  of  the  visible  portions  of  the  surface 
on  which  no  contour  lines  are  shown.  These  portions  may 
indeed  have  any  sinuous  configuration,  though  experience  and 
association  lead  us  to  assume  that  the  figure  represents  an  ellip- 
soidal body,  or  something  like  one. 

100.  We  therefore  continue  the  examination  of  the  action  of 
light  upon  surfaces,  in  order  to  discover  how  their  forms  may  be 
distinguished,  and  we  find  that  a  surface  is  illuminated  in  pro- 
portion to  the  directness  with  which  the  light  falls  upon  it.     See 
PI.  XV.,  Fig.  56.     Here,  let  AB  be  the  trace  of  a  plane,  tangent 
at  T,  to  any  curved  surface.     The  space  ab  is  illuminated  by  the 
beam  of  rays,  M.     The  equal  space  cd,  struck  more  obliquely 
by  the   light,  is  illuminated  by  the  thinner  beam  N,   while 
another  equal'space,  ef,  which  is  struck  perpendicularly,  receives 
all  the  light  of  the  thickest  beam,  O.     Hence,  in  this  case,  the 
curved  surface  is  most  highly  illuminated  at  the  point  of  tan- 
gency,  T.     Along  the  curve  of  shade,  tangent  planes  coincide 
with  the  direction  of  the  light,  and  therefore  the  given  surface 
there  receives  no  light.     The  curve  of  shade  therefore  is  the 
darkest  part  of  the  surface. 

101.  Between  the  curve  of  shade  and   the  brightest   point, 
curves  may  be  conceived  lying  on  the  surface,  at  all  points  "of 
any  one  of  which,  tangent  planes  will  make  equal  angles  with  the 
rays  of  light. 

But  these  angles  represent  the  angles  made  by  the  surface 
itself  with  the  light  along  the  supposed  curve.  Therefore  the 
curves  just  supposed  will  be  curves  of  equal  illumination,  and 
the  illumination  of  the  surface  will  vary  in  intensity  from  the 
maximum  darkness  at  the  curve  of  shade,  where  the  surface 
makes  an  angle  of  0°  with  the  light,  to  the  maximum  brightness, 
where  the  sifrface  is  normal  to  the  light. 

Moreover,  drawing  aS,  for  example,  the  perpendicular  width 
of  the  beam  of  light,  it  represents  the  sine  of  the  angle  SJcz,  made 
by  the  light  with  the  tangent  plane  AB.  Hence,  in  this  view, 
the  intensity  with  which  a  surface  is  illuminated  is  directly  as  the 
sine  of  the  angle  made  by  the  light  with  that  surface. 

102.  But  again :   Bodies   are  not  seen  directly  by  the  light 
thrown  upon  them,  but  by  such  portion  of  that  light  as  is 
returned  from  them. 


SHADES  AND  SHADOWS.  115 

Hence,  in  general  terms,  the  comparative  amounts  of  light  re- 
mitted to  the  eye,  from  the  different  points  of  a  body,  are,  with 
its  apparent  contour,  the  means  of  judging  of  its  form. 

103.  Now,  optical  researches  show,  that,  when  light  falls  upon. 
a  body,  a  part  is  extinguished,  or  destroyed  (absorbed  according 
to  the  material  theory),  a  part  is  reflected,  and  a  third  portion  is 
polarised,  by  refraction  among  the  molecules  of  the  surface. 

104.  Strictly  speaking,  a  body  is  visible,  only  as  it  shows  its 
own  proper  color,  which  it  does  by  means  of  the  refracted  rays 
which  it  remits. 

Mirror-like  surfaces  reflect  the  rays  just  as  they  are  received, 
and  present  images  of  all  objects,  from  which  light  proceeds  to 
fall  upon  them.  They  thus  indicate  their  presence  by  their 
effects,  but  are  strictly  invisible,  except  that,  in  practice,  there 
are  no  absolutely  perfect  mirrors ;  hence  they  are  faintly  visible 
by  the  few  refracted  rays  coming  from  the  molecules  of  their 
surfaces. 

]  05.  The  law  of  reflection  is,  that  the  incident  and  reflected 
rays,  at  the  same  point  of  a  surface,  make  equal  angles  with  that 
surface. 

Hence,  as  there  will,  taking  the  general  case  of  a  double-curved 
surface,  be  but  one  such  point,  for  a  given  fixed  position  of  the 
luminous  point,  and  point  of  sight,  a  perfectly  polished  double- 
curved  body,  exposed  to  a  single  source  of  light,  would  have  but 
a  single  visible  point ;  and  that,  the  one  at  which  a  normal  line, 
or  a  tangent  line  or  plane  to  the  surface,  would  make  equal  angles 
with  the  incident  ray,  and  the  ray  reflected  to  the  eye. 

The  condition  for  the  complete  visibility  of  a  perfectly  polished 
body,  therefore,  is,  that  it  shall  be  exposed  to  light  from  all 
sources. 

106.  The  majority  of  the  objects  which  present  themselves  to 
our  observation,  as  a  stone  or  piece  of  wood,  are  dull,  or  partially 
polished  bodies,  that  is  to  say,  their  exterior,  without  failing  to 
present  an  appearance  of  continuity,  is  nevertheless,  by  their 
porosity,  broken  up  into  minute  asperities  and  cavities,  as 
shown,  greatly  magnified,  in  PI.  XV.,  Fig.  57.  These  irregulari- 
ties, though  separately  invisible,  from  their  minuteness  as  com- 
pared with  the  size  of  the  body,  yet,  under  the  action  of  the  light, 
produce  an  aggregate  effect,  which  is  appreciable,  since  they  are 
of  considerable  magnitude,  as  compared  with  the  extreme  tenuity 
of  the  molecules  of  light.  Each  asperity,  therefore,  is  considered 


116  BRILLIANT  POINTS. 

as  having  a  number  of  indefinitely  small  facets,  one  or  more  of 
which  is  in  a  position  to  return  light  to  the  eye. 

But  the  exterior  facets,  lying  in  the  perfectly  continuous  ima- 
ginary surface  of  the  body,  are  larger  than  interior  ones,  which 
are  withdrawn  from  polishing  agencies ;  and  are  nearer  each  other, 
as  seen  in  projection,  in  the  vicinity  of  the  brilliant  point,  L,  than 
in  the  more  directly  viewed  portion  of  the  continuous  ideal  sur- 
face containing  them ;  and,  besides,  the  properly  disposed  interior 
facets  may  often,  in  obliquely  viewed  regions,  be  hidden  by 
asperities  in  front  of  them,  so  that  the  point  on  a  dull  body, 
which  corresponds  to  the  only  visible  point  of  a  perfectly 
polished  body,  is  the  brightest  point  of  that  dull  body. 

107.  Taking  into  the  account  the  secondary  remissions  to  the 
eye,  from  facets  which  are  illuminated  by  reflection  from  other 
facets,  it  appears  that  the  whole  illuminated  part  of  a  dull  body 
is  rendered  visible  by  a  single  source  of  light. 

Its  shade,  however,  can  become  visible  only  by  sending  to  the 
eye  the  light  received  from  a  secondary  source,  as  from  the 
atmosphere,  or  from  surrounding  bodies. 

Hence,  if  PI.  XV.,  Fig.  55,  represents  a  dull  body,  illuminated 
by  a  single  source  of  light,  all  that  part,  B — FGrDH,  of  the  illu- 
minated part,  which  is  in  the  field  of  vision,  will  be  actually 
visible.  The  completely  unilluminated  shade,  GCDH,  though 
in  the  range  of  vision,  will  be  invisible,  since  it  remits  no  rays  to 
the  eye. 

108.  But  the  theory  of  minute  reflecting  facets  alone,  is  insuf- 
ficient, in  not  accounting  for  the  colors  of  bodies ;  since  light  that 
is  merely  reflected,  retains  the  color  which  it  had  when  coming 
in  incident  rays.     It  has  therefore  been  presumed,  and  is  con- 
firmed by  experiments,  that  bodies  are  visible,  mainly  in  conse- 
quence of  light  remitted  by  them  after  polarization  by  refraction, 
due  to  vibrations  in  their  superficial  molecules.     These  vibrations 
are  supposed  to  be  in  unison — so  to  speak — with  those  of  light 
of  the  color  presented  by  those  molecules ;  and  they  make  those 
molecules  act  as  self-luminous  points  while  illuminated. 

109.  Hence,  besides  the  contours  furnished  by  their  projec- 
tions, the  conditions  for  the  full  representation  of  the  continuity 
and  forms  of  bodies,  are,  that  their  surfaces,  being  dull  or  porous, 
shall  be  illuminated  from  a  primary  and  strong,  and  a  lesser  or 
reflected  light ;  that  their  numberless  facets  shall  be  so  distributed 
as  to  reflect  increasing  quantities  of  light,  in  successive  rings  of 


SHADES  AND  SHADOWS.  117 

equal  reflection,  from  the  curve  of  shade  to  the  brilliant  point, 
analogous  to  the  curves  of  equal  illumination  (101)  and  that  the 
vibrations  of  their  superficial  molecules,  under  the  action  of  light, 
shall  cause  them  to  act,  while  exposed  to  the  light,  as  self-lumi- 
nous bodies. 

Finally :  a  true  representation  of  these  apparent  varying 
intensities  of  light  and  shade,  exhibits  graphically  to  the  eye  the 
continuity  and  the  consecutive  changes  of  form  of  surfaces,  hence, 
in  connection  with  the  apparent  contours  of  bodies,  it  adequately 
represents  them  to  the  eye. 

§  3. — Of  Brilliant  Points  and  Lines. 

110.  Most  conspicuous  and  important,  next  to  the  line  of 
shade  of  a  body,  is  its  brilliant  point,  or  the  element  of  greatest 
apparent  illumination  in  its  illuminated  part.     As  these  bril- 
liant points  can,  moreover,  usually  be  constructed  without  diffi- 
culty, they  will  now  be  more  fully  considered. 

In  treating  of  brilliant  points  thus  in  detail,  the  relative  posi- 
tions of  the  luminous  point,  and  the  point  of  sight,  with  respect 
to  a  given  body,  are  first  to  be  noted.  Since  either  of  these 
points  may  be  either  at  a  finite  or  an  infinite  distance  from  the 
given  body,  the  four  following  combinations  may  arise : 

Distance  of  the 

Luminous  point.  Point  of  sight. 

Infinite.  Finite. 

Finite.  " 

Infinite.  Infinite. 

Finite.  " 

111.  The  two  former  cases  are  properly  treated  under  the 
head  of  "scenographic  projections"  or  natural  perspective,  since 
the  point  of  sight  is  there  supposed  to  be  at  a  finite  distance  from 
the  object  viewed. 

The  two  latter  cases  may  here  be  discussed,  since,  whatever 
the  distance  of  the  source  of  light,  the  point  of  sight  is  at  an 
infinite  distance  from  the  object,  which  is  characteristic  of  ortho- 
graphic projections. 

112.  In  speaking  now  of  the  position  of  the  brilliant  point 
upon  any  surface,  the  distinction  between  the  real,  and   the 


118  BRILLIANT   POINTS. 

apparent  brilliant  point  must  be  observed.  The  real  brilliant 
point  on  a  given  surface,  is  the  point  at  which  the  tangent  plane 
to  that  surface  is  perpendicular  to  the  direction  of  the  light,  for 
this  point  is  in  a  position  to  receive  the  greatest  amount  of  light 
per  unit  of  surface. 

Hence,  if  T,  PL  XV.,  Fig.  56,  be  the  point  of  contact  of  a 
surface  with  a  tangent  plane,  AB,  to  which  the  light  is  normal, 
T  will  be  the  real  brilliant  point  of  the  given  surface. 

113.  The  apparent  brilliant  point  of  a  surface,  is  that  point  at 
•which  a  normal  to  the  surface  bisects  the  angle  between  the 
incident  and  the  reflected  rays  at  the  same  point  (105).     The 
apparent  brilliant  point  is  the  only  one  which  it  is  necessary  to 
consider,  in  treating  of  the  finished  execution  of  shading. 

114.  On  developable  single  curved  surfaces,   and  with    the 
luminous  point  and  point  of  sight  both  at  an  infinite  distance, 
the  brilliant  point  expands  into  a  brilliant  line,  whose  location 
will  be  explained  in  detail,  in  connection  with  the  problems  of 
the  next  section. 


SECTION  II. 
The    Construction    of  Brilliant    3?oints    and.    Lines. 

§  1. — Brilliant  elements  ff  Planes. 

115.  A  plane  illuminated  by  parallel  rays  would  be  equally 
light  in  every  part,  and,  neglecting  the  effect  of  the  different 
depths  of  atmosphere,  through  which  its  different  parts  would 
be  seen  were  it  viewed  obliquely,  every  part  of  it  would  appear 
equally  bright. 

This  result  would  follow  from  the  theory  of  the  constitution 
of  surfaces,  explained  in  connection  with  PI.  XV.,  Fig.  57,  for 
then  a  material  plane  surface  of  uniform  texture  would  present 
a  uniform  distribution  of  facets,  so  disposed  as  to  reflect  rays  to 
the  eye. 

A  plane,  illuminated  by  diverging  rays,  will  have  a  brilliant 
point. 


SHADES  AND  SHADOWS.  119 


PROBLEM  XLII. 

To  find  the  brilliant  point  of  a  plane  which  receives  light  from 
a  near  luminous  point. 

PL  XV.,  Fig.  58.  Let  the  plane  be  vertical,  and  let  TP  be 
its  horizontal  trace  ;  and  let  S  be  the  luminous  point.  Also  let 
TE  be  the  direction  of  the  reflected  rays,  -which  are  parallel  (111). 

At  any  point,  T,  construct  a  line,  TN,  perpendicular  to  the 
given  plane,  and  make  the  angle  KTN=RTN.  Then,  through 
the  luminous  point,  S,  draw  a  ray,  SB,  parallel  to  KT,  and  B, 
its  intersection  with  the  plane  TP,  will  evidently  be  the  point  at 
which  the  incident  and  reflected  rays  make  equal  angles  with  the 
normal  to  the  plane.  Hence  B  is  the  brilliant  point  required. 

Remarks. — a.  Observe,  that  as  NT  is  normal  to  the  given 
plane,  the  plane  of  the  incident  and  reflected  rays  is  perpendicu- 
lar to  the  given  plane.  Hence,  in  this  case,  the  ray  SB  is  hori- 
zontal in  space. 

b.  The  construction  of  the  vertical  projection  of  the  figure, 
and  the  solution  of  the  problem  when  the  given  plane  is  oblique 
to  the  planes  of  projection,  may  now  be  left  to  the  student. 

§  2. — Brilliant  elements  on  Developable  Surfaces. 

Passing  to  single  curved  surfaces,  we  shall  first  consider  deve- 
lopable  surfaces — and  these,  at  first,  as  illuminated  by  parallel 
rays. 

PROBLEM  XLIII. 

To  find  the  brilliant  element  on  a  cylinder,  illuminated  by 
parallel  rays. 

First  Solution.— PL  XV.,  Fig.  59.  Let  ANB— A'B'  be  a 
vertical  right  cylinder,  and  RO — R'O'  the  projections  of  a  ray 
of  light.  OE  represents  the  direction  of  those  reflected  rays 
which  reach  the  eye.  Now  revolve  the  plane,  R'O'E,  of  the 
incident  and  reflected  ray?,  RO — R'O'  and  O' — OE,  about  its 
horizontal  trace,  O'E,  into  the  horizontal  plane  of  projection. 
The  incident  ray  will  then  appear  at  R"0'— R'"O,  and  L'"O  is 


120  BRILLIANT  POINTS. 

the  revolved  position  of  the  bisecting  line  of  the  angle  EOB 
(105).  Then,  by  making  the  counter  revolution,  this  bisecting 
line  will  appear  at  LO — I/O',  since  it  is  in  a  plane  B/O'E  which 
is  perpendicular  to  the  vertical  plane  of  projection. 

Now  we  may  suppose  that  a  row  of  asperities  is  ranged  along 
the  element  N — N'N"  in  which  the  bisecting  line  pierces  the 
cylinder.  If  then  each  of  these  asperities  has  one  or  more 
minute  facets  which  are  perpendicular  to  LO — I/O',  they  will 
collectively  reflect  the  rays  contained  in  a  vertical  plane  of  rays 
through  EO— R'O',  and  thus  form  a  brilliant  line  N— N'N". 

Second  Solution. — In  this  solution,  some  of  the  facets  of  the 
little  asperities  are  supposed  to  be  arranged  in  parallel  elliptical 
bands,  indefinitely  narrow,  and  contained  in  the  successive  paral- 
lel planes  of  incident  and  reflected  rays. 

See  PL  XV.,  Fig.  60,  where  the  light  is  taken  in  the  same 
direction  as  in  the  last  figure,  for  the  sake  of  easy  comparison 
of  the  two.  Those  features  of  the  construction,  which  are  the 
same  in  both  figures,  are  here  omitted. 

HO  is  the  revolved  position  of  an  incident  ray,  OE  is  a 
reflected  ray,  and  DO  is  the  bisecting  line  of  their  included  angle. 
A'B'  is  the  vertical  trace  of  a  plane  of  rays,  perpendicular  to  the 
vertical  plane  of  projection,  and  determining  the  narrow  elliptical 
band  A'B',  along  which  facets  are  supposed  to  be  disposed  so  aa 
to  reflect  light  to  the  eye.  A"CB"S  is  the  revolved  position  of 
this  elliptical  band.  On,  the  bisecting  line  of  parallel  chords,  aa 
C<?  and  #5,  which  are  perpendicular  to  DO,  determines  N'",  the 
point  at  which  a  normal,  parallel  to  DO,  can  be  drawn. 

In  the  counter  revolution,  N'"  returns  to  N;  hence,  if  the  con- 
vex surface  of  the  cylinder  be  supposed  to  be  formed  of  consecu- 
tive bands  of  reflecting  facets  parallel  to  A'B,  a  vertical  row  of 
them  will  ba  found  on  an  element  N — N'N". 

Remarks. — a.  It  should  be  remembered,  that  the  plane  of  the 
incident  and  reflected  ra}7s  is  normal  to  the  reflecting  surface. 

Otherwise,  a  ray  striking  a  plane,  at  a  point  A,  for  example, 
might  be  reflected  in  a  cone  of  reflected  rays,  generated  by  the 
revolution  of  the  incident  ray  about  a  perpendicular  to  the  plane 
at  the  point  A. 

But  in  the  case  of  an  absolutely  nnporous  and  polished  cylin- 
der, or  cone,  which  is  struck  obliquely  by  the  light,  a  plane  of 
incident  and  reflected  rays,  situated  as  in  the  present  example, 
cannot  be  made  normal  to  the  convex  surface  at  N.  Hence  such  a 


SHADES  AND  SHADOWS.  121 

cylinder,  when  illuminated  from  a  single  source  of  light,  should 
be  absolutely  invisible  throughout. 

The  existence  of  asperities  of  surface,  presenting  reflecting 
facets,  is,  therefore,  a  necessary  condition  for  the  visibility  of  the 
surface  thus  illuminated.  Hence  reasoning  or  experiment  must 
decide  which  of  the  theories  of  the  location  of  the  reflecting 
facets,  given*  in  the  two  solutions  of  this  problem,  is  true,  since 
these  solutions  give  quite  different  locations  to  the  brilliant  line. 

The  first  construction  is,  I  believe,  the  only  one  hitherto  used, 
excepting  the  approximate  one  given  in.  my  "  Elementary  Pro- 
jection Drawing,"  and  it  agrees  with  the  formation  of  the  artifi- 
cial grain  of  the  surface  by  the  operation  of  turning. 

b.  Either  of  the  solutions  of  this  problem  may  be  applied,  by 
the  student,  in  finding  the  brilliant  element  of  a  cone. 


PROBLEM  XLIV. 

To  find  the  brilliant  point  of  a  cylinder  which  is  illuminated 
by  diverging  rays. 

PI.  XV.,  Fig.  61.  Let  L  be  the  luminous  point,  and  BDK  the 
horizontal  projection  of  the  cylinder.  The  reflected  rays  being 
parallel,  because  the  eye  is  at  an  infinite  distance  from  the 
cylinder,  let  LG  be  a  line  in  the  direction  of  the  reflected  rays. 

All  lines  through  O,  and  in  a  plane  perpendicular  to  the  axis 
of  the  cylinder,  will  be  normal  to  its  surface.  The  cylinder  being 
vertical,  in  the  present  case,  this  plane  will  be  horizontal,  and 
EO,  FO,  GO,  etc.,  are  normals  to  the  cylinder. 

Now,  making  La=LE  ;  LJ=LF,  etc.,  the  curve  abc  is  the  line 
at  each  of  whose  points  incident  rays,  as  L«,  and  reflected  rays 
from  a,  etc.,  parallel  to  LG,  make  equal  angles  with  the  normals, 
as  E#O,  etc.,  through  the  same  points. 

Hence  B,  where  this  curve  meets  the  convex  surface  of  the 
cylinder,  is  the  point  of  that  surface  at  which  the  incident  and 
reflected  rays,  LB  and  BY,  make  equal  angles  with  the  normal 
BO.  That  is,  B  is  the  horizontal  projection  of  the  required  bril- 
liant point.  Its  vertical  projection  is  not  shown,  it  being  found 
by  merely  projecting  B  into  the  circle  cut  from  the  cylinder  by  a 
horizontal  plane  through  the  luminous  point. 


122  BRILLIANT   POINTS. 


PROBLEM  XLY. 

To  find  the  brilliant  point  on  a  cone  which  is  illuminated 
from,  a  near  luminous  point. 

PI.  XY.,  Fig.  62.  In  the  last  problem,  the  curve  in  which  the 
normals  pierced  the  cylinder,  coincided,  in  horizontal  projection, 
with  the  projection  BDK  of  the  cylinder,  and  was,  therefore,  not 
constructed.  The  cone,  not  having  a  vertical  surface,  this  curve 
of  intersection  of  the  normals  would  appear  as  a  distinct  curve, 
which  must  be  constructed. 

Let  VDTH — Y'D'H  be  a  cone  of  revolution,  having  its  axis 
vertical;  and  let  S,A'  be  the  luminous  point. 

Draw  SAi",  in  the  direction  of  the  reflected  rays.  At  S,  any 
line  will  make  equal  angles  with  S5",  and  some  incident  ray, 
hence  S  is  a  point  of  the  trial  curve,  containing  the  intersection 
of  normals  with  incident  rays.  A  is  another  point  of  the  same 
curve,  since  SA,  regarded  first  as  an  incident,  and  then  as  a 
reflected  ray,  makes  the  same  angle  with  a  normal  YA  at  right 
angles  to  it. 

To  find  other  points  of  the  trial  curve  SAA.  At  any  point,  as 
5",  on  the  line  S5",  draw  the  horizontal  projection,  b"V,  of  a 
normal.  Revolve  this  normal  to  the  position  Z»Y,  parallel  to  the 
vertical  plane  of  projection.  N'J',  perpendicular  to  the  element 
V'D' — taken  as  the  revolved  position  of  the  element  Yy — will 
then  be  its  vertical  projection,  and  A'N'  is  the  vertical  projection 
(not  drawn)  of  its  primitive  position.  Now  revolve  this  normal 
and  the  line  SJ",  about  an  axis  perpendicular  to  the  vertical  plane 
at  N',  and  into  the  horizontal  plane  N'A".  It  will  then  appear 
at  A"N'— BY,  and  S&"  will  appear  at  A"— S'"B.  Now  make 
S'"&"— not  drawn— equal  to  S"'B,  and  h"  will  be  the  revolved 
position  of  another  point  of  the  trial  curve.  In  the  counter- 
revolution, h"  returns  in  a  short  arc,  h"h,  parallel  to  Bi",  and  ^, 
on  the  primitive  projection,  b"V,  of  the  normal,  is  another  point 
of  the  curve  S  AA. 

Having  now  found  one  point  of  this  curve,  in  the  manner  just 
described,  others  are  easily  found  as  follows,  Revolve  assumed 
points,  asp  and  #,  top"'  and  #,  and  join  p"  and  k  with  Y.  Then 
arcs,  with  radii  S'"p"  and  S'"&,  will  locate  points  on  p"V  and 
&Y — not  shown — at  distances  from  S"',  equal  to  S'"pr/  and  S"'&. 


SHADES  AND  SHADOWS.  123 

Then,  by  counter-revolution,  as  before,  find  c  and  y.  The  trial 
curve  SAcA  can  now  be  drawn. 

It  is  necessary,  in  the  next  place,  to  construct  the  curve  in 
•which  all  normals,  which  intersect  SJ",  pierce  the  conic  surface. 
Take,  for  example,  the  normal  Y5".  After  revolution  into  the 
meridian  plane  AY,  parallel  to  the  vertical  plane,  this  normal 
appears  at  Vb — JV",  normal  to  the  revolved  position,  Y'D',  of 
the  element  Vr,  which  is  in  the  meridian  plane  through  Y&". 
Hence  r'",r"  is  the  revolved,  and  r,/  the  primitive  position  of 
the  intersection  of  this  normal  with  the  conic  surface. 

Other  similar  points  being  found  in  like  manner,  the  curve 
uer — e'n'r'  may  be  sketched  as  the  locus  of  the  intersections  of  all 
normals,  through  points  of  A' — S6",  with  the  conic  surface.  This 
curve  intersects  SA^,  the  locus  of  the  intersections  of  normals 
with  incident  rays,  at  n,n'.  This  point  is  therefore  that  point,  on 
the  conic  surface,  at  which  the  incident  and  reflected  rays,  Sn 
and  wE,  make  equal  angles  with  the  normal,  wY,  at  that  point. 
Note,  however,  that  these  angles  are  not  equal  in  projection,  since 
their  plane  is  oblique. 

Remark, — "We  may  construct  the  point  in  which  the  normal  at 
any  point  will  pierce  the  line  Sb".  Thus,  let  F  be  the  revolved 
position  of  any  point  at  which  a  normal  is  to  be  drawn.  FK, 
perpendicular  to  Y'H',  is  the  revolved  position  of  this  normal. 
The  revolution  of  FK,  about  Y'K  as  an  axis,  will  generate  a 
cone,  whose  surface  will  be  normal  to  that  of  the  given  cone ; 
then  the  intersection  of  this  auxiliary  cone  with  the  line  Sb" — A', 
will  give  the  point,  which,  when  joined  with  Y,K,  will  be  the 
normal  required. 


§  3. — Brilliant  Points  on  Warped  Surfaces. 
PROBLEM  XLYI. 

To  find  the  brilliant  point  of  a  warped  surface,  when  illuminated  ~by 

parallel  rays. 

The  solution  of  this  problem  involves  the  construction  of  a 
normal,  parallel  to  a  given  line,  L,  viz.,  to  the  bisecting  line  of 
the  angle  included  by  an  incident  and  a  reflected  ray. 

But  since  these  rays  make  equal  angles  with  a  tangent  plane 


124  BRILLIANT  POINTS. 

which  is  perpendicular  to  the  supposed  normal,  the  construction 
of  the  normal  may  be  effected  by  the  construction  of  a  tangent 
plane,  perpendicular  to  a  given  line.  But,  again,  since  all  planes 
which  are  perpendicular  to  the  same  straight  line,  are  parallel, 
the  operation,  just  proposed,  is  equivalent  to  the  construction  of 
a  tangent  plane  parallel  to  a  given  plane.  • 

To  construct  the  tangent  plane,  draw  any  two  lines,  A  and  B, 
not  parallel,  but  each  perpendicular  to  the  given  line,  L.  A 
system  of  tangent  lines,  parallel  to  A,  will  constitute  a  cylinder, 
tangent  to  the  warped  surface.  Another  such  system,  parallel  to 
B,  will  form  another  tangent  cylinder.  The  intersection  of  the 
curves  of  contact  of  these  cylinders  will  be  the  point  of  taiigency, 
on  the  warped  surface,  of  a  tangent  plane  perpendicular  to  L,  and 
therefore  the  point  of  intersection,  I,  of  the  required  normal, 
parallel  to  L.  Hence  I  will  be  the  required  brilliant  point. 


PROBLEM  XLYII. 

To  construct  the  brilliant  point  on  a  screw,  when  illuminated  by 

parallel  rays. 

Produce,  each  way,  several  elements  on  each  side  of  the  meri- 
dian plane  parallel  to  the  vertical  plane,  the  axis  of  the  screw 
being  vertical,  so  as  to  draw,  tangent  to  them,  a  sufficient  arc  of 
that  meridian  curve  of  the  screw,  which  lies  in  this  meridian 
plane.  Then,  as  in  Prob.  XLL,  find  the  bisecting  line,  N,  of  the 
angle  included  between  an  incident  and  a  reflected  ray.  Revolve 
the  line,  N,  till  parallel  to  the  vertical  plane  of  projection.  Draw 
a  line,  parallel  to  this  revolved  position,  and  normal  to  the  meri- 
dian curve,  just  named,  at  a  point,  n.  This  point,  n,  will  be  the 
revolved  position  of  the  brilliant  point.  In  counter-revolution, 
it  will  return  in  a  helical  arc,  and  its  projections  will  appear  on 
the  projections  of  N. 

This  construction  can  be  wrought  out  by  the  student. 


SHADES  AND  SHADOWS.  125 

§  4. — Brilliant  Points  on  Double-  Curved  Surfaces. 
PROBLEM  XLVIII. 

To  find  tfie  brilliant  point  on  any  double-curved  surface,  when 
illuminated  by  diverging  rays. 

Let  PI.  XV.,  Fig.  63,  represent  any  double-curved  surface, 
not  of  revolution.  Let  S  be  the  luminous  point,  and  SB/  a  line 
parallel  to  the  reflected  rays.  Let  n'Nn"  be  the  curve  in  which 
all  normals,  as  RW,  RN,  and  R'W,  intersect  the  given  surface. 
Also,  let  N'NN"  be  the  locus  of  intersections  of  these  normals 
with  incident  rays  from  S.  That  is,  SN'=SR',  SN=SR,  and 
SN"=SR",  etc.,  so  that  if  the  rays,  SN',  etc.,  were  drawn,  the 
angles  at  N',  etc.,  made  by  the  incident  ray  SN',  etc.,  and  normals 
R'N',  etc.,  would  be  equal  to  the  angles  at  R',  etc.,  made  by  the 
reflected  ray  SR'  with  the  same  normals.  Then  N,  the  intersec- 
tion of  the  two  curves  just  described,  is  that  point  on  the  given 
surface,  where  the  incident  and  reflected  rays  make  equal  angles 
with  the  normal  RN.  Hence  N  is  the  brilliant  point. 

The  practical  difficulty,  in  case  of  the  class  of  surfaces  con- 
sidered in  this  article,  and  also  in  case  of  warped  surfaces,  is,  that, 
unless  they  are  surfaces  of  revolution,  there  is  no  convenient 
direct  construction  of  the  required  normals.  Hence  the  remain- 
ing problems  embrace  only  the  construction  of  the  brilliant  points 
of  double-curved  surfaces  of  revolution,  when  illuminated  by 
parallel  rays. 

"When  such  surfaces  are  exposed  to  diverging  rays,  their 
brilliant  points  are  found  as  in  Prob.  XLIY. 

PKOBLEM  XLIX. 
To  find  the  brilliant  point  on  a  sphere,  illuminated  by  parallel  rays. 

PI.  IX.,  Fig.  26.  A  vertical  plane,  through  the  centre  of  the 
sphere,  is  here  regarded  as  the  vertical  plane  of  projection.  OL 
is  the  vertical  projection  of  a  ray  through  the  centre  of  the  sphere. 
As  no  horizontal  projection  of  the  sphere  is  shown,  Or  may  be 
assumed  as  the  ray,  after  being  revolved  into  the  vertical  plane 


126  BRILLIANT    POINTS. 

of  projection,  about  LK,  a  line  of  that  plane,  as  an  axis.  Then, 
also,  OP  will  represent  the  direction  of  the  reflected  rays.  Hence 
N'O,  the  revolved  position  of  the  bisecting  normal  of  the  angle 
rOP,  gives  N'  as  the  revolved  position  of  the  brilliant  point. 
By  a  counter-revolution  about  LK,  its  primitive  position,  N,  ia 
found. 

PROBLEM  L. 
To  find  the  brilliant  point  on  a  piedouche. 

PI.  X.,  Fig.  31.  A^ — SY  is  any  incident  ray  intersecting  the 
axis  A — T'A'.  Let  S' — AW  represent  the  direction  of  the 
reflected  rays,  as  the  piedouche  is  seen  in  vertical  projection. 
Revolving  the  assumed  ray  around  AW,  as  an  axis,  till  it  is 
parallel  to  the  horizontal  plane  of  projection,  it  appears  at  SY" — 
Ai".  Then  draw  Ag-,  the  bisecting  line  of  the  angle  t'"AW, 
and  draw  an  auxiliary  line,  i"W.  In  the  counter-revolution, 
i"j!"  returns  to  ij!,  the  line  i"W  returns  to  i W,  and  q  to  p, 
whence  it  is  projected  to  p',  because  the  plane  WAi  is  perpen- 
dicular to  the  vertical  plane  of  projection.  Now  pA  and  p'S' 
are  the  projections  of  the  bisecting  line  of  the  angle  iAW.  The 
brilliant  point  is  the  point  of  contact  of  a  tangent  plane  which 
is  perpendicular  to  this  bisecting  line.  Then  revolve  Ap — S'jp', 
till  parallel  to  the  vertical  plane,  taking  A — T'A'  for  an  axis, 
when  it  will  appear  at  Ap" — S'p"'.  Next,  draw  the  normal  fl, 
parallel  to  ^'''S',  and  I  will  be  the  revolved  position  of  the  bril- 
liant point.  Construct  the  projections  of  the  horizontal  circle 
through  this  point,  and  V,V,  its  intersection  \\ith  the  meridian 
plane-  through  the  bisecting  line  pA — p'A,  is  the  primitive 
position  of  the  required  brilliant  point. 

Remark.  The  student  may  now  construct  the  brilliant  point 
on  any  of  the  double  curved  surfaces  of  revolution  shown  in  the 
preceding  plates,  or  upon  a  warped  hyperboloid  of  revolution. 


SHADES   AND  SHADOWS.  12 T 


CHAPTER  H. 

OP  THE  REPRESENTATION  OF  THE  GRADATIONS 
OF  LIGHT  AND  SHADE. 

116.  In  observing  a  shade,  or  a  shadow,  two  things  arrest 
the  attention,  its  position,  and  its  intensity. 

To  find  the  position  of  shades  and  of  shadows  upon  surfaces, 
was  the  object  sought  in  "  BOOK  I."  Now,  we  are  to  determine 
their  varying  intensities,  as  affected  by  the  several  circumstances 
presently  to  be  enumerated.  Furthermore,  the  varying  apparent 
intensity  of  the  illumination  of  those  parts  of  surfaces  which 
are  exposed  to  the  light,  will  be  affected  by  the  same  circum- 
stances. Hence  the  gradations  of  the  lights,  and  of  the  shades 
and  shadows  of  surfaces,  will  be  discussed  together. 

117.  The  particulars  affecting  apparent  intensities,  and  the 
gradations  of  lights  and  shades,  and  which  will  next  be  dis- 
cussed in  detail,  are  arranged  under  the  following  heads  : 

1°. — Effects  due  to  the  laws  of  illumination  and  vision. 

2°. — Those  due  to  the  supposed  infinite  distances  of  objects 
from  the  eye. 

3°. — Those  due  to  secondary  sources  of  light;  including  the 
air  as  an  illuminating  medium. 

4°. — Those  due  to  the  nature  of  different  surfaces. 

5°. — Those  due  to  the  atmosphere  as  an  absorbing  medium. 

6°. — Those  due  to  variations  in  the  intensity  of  the  light. 

7°. — Those  due  to  the  forms  of  bodies. 

8°. — Those  due  (on  the  drawing)  to  drawing  materials  and 
processes  of  manipulation. 

118.  Effects  due  to  the  laws  of  illumination  and  vision. 

a.  Law  of  illumination.  The  intensity  of  light,  or  of  the 
degree  of  actual  illumination  of  the  same  surface,  at  different 
distances  from  the  source  of  light,  varies  inversely  as  the  square 
of  the  distance  from  the  luminous  source. 


128  THEORY  OF  SHADING. 

b.  Case  where,  the  light  comes  from  an  infinite,  distance.     The 
principle  just  stated  applies,  sensibly,  whenever  the  source  of 
light  is  at  a  finite  distance.     But  when  the  light  proceeds  from 
an  infinite  distance,  terrestrial  bodies  are  all  at  substantially  the 
same  distance  from  it,  and  when  similarly  situated  with  respect 
to  it,  may  be  considered  as  equally  illuminated. 

c.  Law  of  apparent  brilliancy.     From  remark  a  it  follows, 
that  if  a  surface  be  illuminated  in  a  given  degree,  its  apparent 
brilliancy  should  be  inversely  as  its  distance  from  the  eye,  but 
its  image  being  reduced  in  the  same  proportion,  the  retina  re- 
ceives light  at  the  same  rate  or  amount  per  superficial  unit  of  its 
surface.     Hence,  practically,  the  apparent  brilliancy  of  a  surface 
of  given  brightness,  and  in  a  given  position,  will  be  the  same  at 
whatever  distance  it    is    seen ;    if  we  disregard   atmospheric 
effects. 

d.  Illustration.     The  last  remark  is  easily  illustrated.     See 
PI.  XV.,  Fig.  64 

Let  A  be  a  visible  point  of  an  illuminated  surface,  and  let  PQ 
be  the  pupil  of  the  eye.  The  point  A  remits  to  the  eye  the  cone 
of  rays  APQ,  whose  base  is  the  pupil  PQ.  These  rays  converge 
in  the  eye,  forming,  on  the  retina  at  R,  the  image  of  the  point 
A.  If  now  the  object  be  removed  to  A',  double  the  distance 
AC,  the  base,  PQ,  remaining  of  constant  size,  the  section  of  the 
new  cone,  at  the  distance  equal  to  AC  from  its  vertex,  will  evi- 
dently have  an  area  equal  to  one-fourth  of  PQ.  But  sections  at 
a  constant  distance  from  their  vertices  measure  the  relative  angu- 
lar capacities  of  the  two  cones.  Hence  the  eye  at  double  the  dis- 
tance AC  receives  one-fourth  as  much  light  from  the  point  A  as  at 
the  distance  AC. 

But  to  apply  this  result  practically,  we  must  consider  more 
than  a  solitary  point.  Two  points  will  be  sufficient.  Then 
let  B  be  a  second  point  on  the  body  AB.  The  axis,  Br,  of  its 
cone  of  rays  gives  r  as  the  focus  of  those  rays  on  the  retina,  and 
Rr  as  the  space  on  the  retina,  illuminated  by  rays  from  the  space 
AB  on  the  object.  Now  remove  AB  to  double  the  distance  AC 
viz.  to  A'B',  and  the  point  B'  will  be  imaged  at  r',  one-fourth  as 
brightly  as  at  r.  But  Rr'  is  half  of  Rr,  and  hence  the  area,  of 
which  Rr'  is  the  diameter,  is  one-fourth  as  large  as  that  whose 
width  is  Rr.  While,  therefore,  A'B'  remits  to  the  eye  one-fourth 
of  the  light  that  AB  remits,  that  light  is  received  by  the  area, 
Rr',  of  the  retina,  which  is  one-fourth  of  the  area  Rr.  Hence  the 


SHADES  AND  SHADOWS.  429 

rate  of  illumination  of  the  retina,  for  a  unit  of  surface,  is  the 
same  in  both  cases. 

e.  In  discussing  this  nice  point,  on  which  learned  authors 
differ,  care  must  be  taken  not  to  confound  aggregate  amount, 
and  consequent  dazzling  effects  of  brilliancy,  with  its  rate  or  inten- 
sity. A  square  foot  of  sun-lightened  snow  is  as  intensely  bright, 
as  to  its  degree  of  illumination,  as  a  whole  snow-bank,  but 'the 
total  amount  of  light  from  it,  and  extent  of  retina  affected  by  it, 
is  less  than  in  case  of  the  whole  snow  drift.  Hence  the  amount 
of  dazzling  effect  is  less.  So  a  star,  as  brilliant  as  the  sun,  can 
easily  be  gazed  upon,  though  its  sensibly  equal  intensity  of  bright- 
ness with  the  sun  is  shown  by  its  twinkling,  which  is  its  daz- 
zling effect  upon  a  mere  point  of  the  retina ;  an  effect  which  has 
only  to  be  repeated  in  the  same  degree  on  many  points  at  once 
of  the  retina,  in  order  to  become  blinding. 

/.  /Shading  of  the  penumbra.  When  a  penumbra  (85)  is  occa- 
sioned, as  it  is,  by  a  near  radiant  body  of  sensible  size,  it  receives 
light  from  a  larger  and  larger  portion  of  the  luminous  body,  as 
it  approaches  its  own  outer  boundary.  Hence  its  shaded  repre- 
sentation would  be  lighter  and  lighter  in  the  same  direction, 
unless  prevented  by  the  form  of  the  surface  containing  it,  or  by 
other  special  circumstances. 

119.  Effects  due  to  the  infinite  distance  of  objects  from  the  eye. 

a.  Objects  at  an  infinite  distance  seen  as  points.  At  the  outset, 
and  as  a  preliminary  remark,  it  should  be  noted  that  supernatu- 
rally  acute  powers  of  vision  are  required  to  perceive  an  object 
at  all,  at  an  infinite  distance.  This  may  be  proved  by  a  refer- 
ence to  the  laws  of  vision,  as  follows : 

Each  molecule  of  the  surface  of  an  illuminated  object,  having 
facets  so  disposed  as  to  diffuse  light  in  all  directions,  remits  to 
the  eye  a  cone  of  rays  whose  base  is  the  pupil  of  the  eye  (11  Sd). 
This  cone  of  rays,  in  passing  through  the  lenses  of  the  eye,  con- 
verges to  a  point  in  its  axis,  and  on  the  retina,  which  is  the 
image  of  the  point  on  the  object  from  which  the  rays  proceeded. 
Each  point  of  the  object  thus  produces  its  image,  and  these 
image  points,  collectively,  form  the  superficial  image  of  the 
object. 

But  now  conceive  the  object  viewed  to  be  at  an  infinite  dis- 
tance from  the  eye.  Frusta,  of  finite  length,  of  all  the  cones  of 
rays  remitted  to  the  eye,  and  having  the  pupil  for  their  common 
base,  will  sensibly  coincide  in  a  single  surface,  which  will  be 

9 


130  THEORY  OF  SHADING. 

sensibly  cylindrical.  These  frusta  of  rays,  having  thus  a  com- 
mon axis,  converge  to  a  single  common  point  on  the  retina,  which 
is  the  image  of  the  object.  Now  in  order  to  be  sensible  of  this 
image,  consisting  of  but  a  single  point,  the  retina  must  be  of 
extraordinary  sensibility — unless  the  light  be  of  extreme  intensity 
— also,  its  nervous  texture  must  possess  absolute  continuity,  in 
order  to  be  impressible,  strictly,  at  every  point. 

b.  Why  represented  as  having  sensible  magnitude. — In  the  two 
respects  just  named,  then,  the  eye  must  possess  extraordinary 
power,  in  order  to  perceive  infinitely  remote  objects ;  but,  since 
they  should  appear  as  points,  why  are  they  represented  as  of 
sensible  magnitude  ?    For  the  same  reason  that  applies  to  ordinary 
vision.     A  giant  nine  feet  high,  and  ninety  feet  distant,  appears 
of  the  same  size,  nominally,  or  in  a  merely  geometrical  sense,  as 
a  boy  three  feet  high,  and  thirty  feet  distant.     But,  practically, 
we  say  that  each  appears  as  we  realize  that  it  appears,  and  the 
appearance  that  we  realize,  i.  e.,  the  one  that  seems  real  to  us,  is 
determined  by  our  knowledge,  derived  from  other  sources,  of  the 
real  sizes  of  objects.     Knowledge,  therefore,  of  the  actual  sizes 
of  objects,  may  make  us  believe  that  an  infinitely  remote  object, 
seen  with  the  organs  above  described,  does  appear  of  sensible 
size,  according  to  that  knowledge  ;  rather  than  as  a  point,  accord- 
ing to  its  minute  image  on  the  retina. 

c.  Theory  of  exaggeration  of  effects  in  shaded  projections.     But 
again,  eyes  of  such  acute  sensibility,  would  also  indicate  with 
corresponding  vividness  the  modifications  of  light  and  shade  due 
to  the  causes  already  stated  (117),  hence  the  usual  practice  of 
greatly  exaggerating  effects,  is  quite  rational  in  shading  objects 
when  shown  in  projection. 

This  exaggeration  appears  principally  in  two  particulars: 
First,  in  the  extremely  vivid  contrast,  between  the  element  of 
shade  and  the  brilliant  point,  which  does  not  appear  in  common 
experience,  but  which  appears  natural,  and  agreeably  suggestive 
of  the  forms  of  objects  when  represented  in  shaded  projections. 
Second,  in  the  effect  allowed  in  representing  surfaces  at  different 
distances.  This  is  shown,  first,  in  the  manner  of  shading  single 
surfaces  seen  obliquely ;  and,  second,  in  the  tinting  of  a  series  of 
parallel  surfaces,  at  slightly  increasing  distances,  with  tints  of 
increasing  darkness. 

120.  Effects  due  to  secondary  sources  of  light;  including  the 
air  as  an  illuminating  medium. 


SHADES  AND   SHADOWS.  131 

a.  Direction  of  the  strongest  secondary  light.     A  surface  will 
diffuse  ligbt  in  quantities  proportioned  to  the  amount  received. 
It  will  receive  most  when  struck  perpendicularly.     Therefore 
the  greatest  intensity  of  light  from  the  particles  of  air,  and  of 
surrounding  surfaces,  will  be  in  a  direction  opposite  to  that  of 
the  primitive  rays. 

b.  Of  the  relative  darkness  of  different  portions  of  the  shade  on  a 
given  surface.     From  (a)  it  follows  that  the  shade  of  a  body  will 
have  its  faint  brilliant  element,  due  to  the  secondary  lights,  just 
as  the  illuminated  part  has  its  own  brilliant  element,  due  to  the 
primary  light.     Therefore  the  shade  is  shaded  lighter,  in  reced- 
ing from  the  line  of  shade,  which  is  the  darkest  line  of  the  body 
in  reference  to  both  lights. 

c.  Of  the  relative  darkness  of  different  parts  of  a  shadow.     Any 
point  in  a  shadow  receives   diffused   light  from   surrounding 
objects,  except  within  a  cone  whose  vertex  is  the  point,  and 
whose  base  is  the  curve  of  contact  of  the  cone  with  the  body 
casting  the  shadow.     Therefore,  generally,  the  further  a  point 
of  shadow  is  from  the  object  casting  it,  the  more  diffused  light 
it  receives,  and  the  lighter  it  is. 

The  general  principle  just  before  named,  must,  however,  be 
applied  by  inspection  to  each  particular  case  of  shadows  on 
plane,  convex,  and  concave  surfaces.  In  case  of  a  vertical  prism 
casting  its  shadow  on  the  horizontal  plane,  the  shadow  would 
be  darkest  at  its  junction  with  the  base  of  the  prism;  a  result 
which  is  confirmed  by  observation. 

d.  Optical,  and  actual  effects  of  near  surrounding  objects.     When 
an  illuminated  surface  is  placed  quite  near  a  shade,  it  actually 
illuminates  a  small  neighboring  region  of  that  shade  to  a  notice- 
able extent,  and  that  region  should  therefore  receive  a  lighter 
tint,  in  a  drawing  showing  both  bodies. 

Besides  this  actual  effect,  there  are  optical  effects,  due  merely  to 
the  mutual  influence  of  contrasts,  as  when  a  shadow  falls  on  an 
illuminated  surface,  and  when  two  tints  of  widely  different  in- 
tensity are  brought  together.  Here,  the  light  surface  appears 
lighter,  and  the  dark  surface  darker,  by  contrast,  as  when  black 
velvet  laid  upon  black  cloth  makes  the  latter  look  lighter. 
Likewise,  complementary  colors  heighten  the  effect  of  each  other, 
when  brought  together.  Thus,  purple  and  gold,  or  blue  and 
orange,  each  look  more  brilliant  when  side  by  side,  than  when 
seen  separately.  Such  contrasts  are  presented  in  the  shaded 


132  THEORY  OF   SHADING. 

drawing  of  an  assemblage  of  objects,  as  well  as  by  the  original 
objects,  hence  the  shading  need  not  be  exaggerated  in  order  to 
produce  the  proper  effect.  Thus,  if  a  shadow  on  a  brilliant 
surface  be  of  nearly  uniform  intensity,  as  it  would  be  on  a 
plane  perpendicular  to  the  light,  its  edges  would,  as  they  should 
do,  appear  darker  than  the  interior  portions,  by  contrast  with  the 
high  light  around  them. 

e.  Of  the  relative  darkness  of  a  shade  and  a  shadow.     Some 
interesting  principles  are  suggested  by  considering  the  relative 
darkness  of  a  shade. and  of  a  shadow,  under  various  circum- 
stances.    For  example,   let  a  shadow  fall  on  the  surface  of  a 
sphere.     Any  point  in  this  shadow,  will  receive  scattering  rays 
of  reflected  and  refracted  light  from  all  illuminated  bodies  and 
particles,  in  the  hemisphere  of  space  bounded  by  a  tangent  plane 
at  the  given  point,  and  exterior  to  the  cone,  whose  vertex  is  the 
same  point,  and  whose  base  is  its  curve  of  contact  with  the  body 
casting  the  shadow.     A  point  in  the  shade  of  the  sphere,  near  to 
which  no  other  surface  receives  a  shadow,  will  be  illuminated  by 
reflections  from  the  entire  hemisphere  of  space,  bounded  by  the 
tangent  plane  at  that  point.     Hence  the  latter  point  would  be 
somewhat  lighter  than  the  former ;  as  also  from  the  fact  of  its 
receiving  the  strongest  atmospheric  reflection  (a). 

f.  In  the  case  of  the  line  of  shade,  as  compared  with  a  shadow 
on  the  illuminated  part  of  a  body,  the  shadow  is  the  darker  of 
the  two.     For,  as  the  atmospheric  reflections  are  strongest  in  a 
direction  opposite  to  the  light,  they  are  weakest  or  nothing  in 
the  same  direction  as  the  light.     Hence  shadows  cast  upon  sur- 
faces in  the  vicinity  of  their  brilliant  points  are  darker  than  the 
lines  of  shade  of  those  surfaces.     The  shadow  of  the  abacus 
upon  the  cylinder,  for  example,  is  darkest  at  the  brilliant  ele- 
ment of  the  cylinder,  and  fades  away  to  the  element  of  shade, 
•where  it  disappears. 

g.  When  the  primary  and  secondary  lights  are  of  nearly  equal 
intensity,  as  seen  in  the  comparatively  uniform  diffusion  of  light 
in  a  cloudy  day,  the  contrast  between  the  brilliant  point,  and  the 
curve  of  shade,  and  the  shadows,  would  be  diminished.  The 
latter  would  appear  lighter,  and  the  former  darker,  than  when 
exposed  to  a  single  strong  light. 

121.  Effects  due  to  the  nature  of  surfaces. 

a.  The  nature  of  surfaces  may  vary,  by  being  dull  or  polished, 
or  by  being  differently  impressed,  in  their  molecules,  by  different 


SHADES  AND  SHADOWS.  133 

colored  rays.  While  no  surfaces  are  of  absolutely  perfect  polish, 
owing  to  their  porosity,  yet,  as  they  approximate  to  such  a 
polish,  it  may  be  possible  to  express  that  fact  graphically  by  the 
manner  of  shading  them. 

A  perfectly  polished  body,  we  have  seen,  has  a  single,  well  de- 
fined brilliant  point.  Hence,  other  things  remaining  the  same, 
the  more  polished  the  surface,  the  smaller  may  be  the  region  of  lifjld 
shading  containing  the  brilliant  point ;  and  the  more  dull  the  surface, 
the  broader  may  be  the  area  of  light  shading — as  if  each  large 
asperity  on  a  dull  body,  which  is  pretty  directly  illuminated, 
had  its  own  somewhat  conspicuous  brilliant  point. 

b.  As  already  stated  (104)  bodies  are,  strictly  speaking,  not 
seen  by  their  reflected  light,  but  by  that  which  they  refract.     Re- 
flected light  merely  furnishes  images  of  the  objects  which  send 
rays  to  the  reflecting  surface,  just  as  a  surface  echoes  sound 
without  imparting  to  it  any  new  quality  of  tone  derived  from 
the  echoing  surface. 

In  accounting  for  the  colors  of  bodies,  an  analogy  is  supposed 
to  exist  between  them  and  musical  instruments,  or  resonant 
volumes  of  air.  Many  persons  may  have  observed  that  upon 
sounding  various  notes,  in  a  clear  and  strong  humming  manner, 
in  a  closed  closet  or  small  room,  some  of  them  will,  without 
extra  effort,  sound  much  louder  than  others,  as  if  the  whole 
body  of  air  in  the  room  were  excited  to  musical  vibrations  in 
unison  with  the  loud  sounding  note.  So  surfaces,  when  exposed 
to  the  pulsations  of  light,  seem  to  respond,  in  the  vibrations  of 
their  molecules,  only  to  the  vibrations  of  rays  of  a  certain  color. 
By  thus  responding,  these  surfaces  become,  for  the  time,  lumi- 
nous bodies,  emitting,  however,  only  such  colored  rays  as  their 
nature  causes  them  to  be  excited  by. 

Now,  as  bodies,  in  proportion  to  the  perfection  of  their  polish, 
reflect  more  and  more  light,  less  will  remain  to  express  their 
color  by  the  refracting  process  just  explained.  Hence  the  higher 
the  polish,  the  less  clearly  will  the  proper  color  of  a  surface  be 
revealed. 

c.  It  may  be  added,  as  a  matter  of  interest,  to  the  illustration 
of  the   closed  room,  that  such  inclosed  spaces  are  resonant, 
though  in  a  less  degree,  to  notes  which  differ  but  little  from  the 
principal  note.     So  it  is  found,  also,  that  the  colors  of  bodies  are 
not  absolutely  pure,  but  are  always  mixed  with  small  portions 
of  the  neighboring  colors  of  the  spectrum.     Thus  the  light  from 


134  THEORY  OF  SHADING. 

a  clear  yellow  flower,  when  dispersed  by  a  prism,  shows  some 
orange  and  green  rays,  mixed  with  the  pure  yellow  ones. 
122.  Effects  due  to  the  atmosphere  as  an  absorbing  medium. 

a.  Appearance  of  an  illuminated  surface  seen  obliquely.     The 
atmosphere,  not  being  perfectly  transparent,  extinguishes  a  por- 
tion of  the  light  coming  through  it,  and  this  portion  is  greater  in 
proportion  to  the  extent  of  air  traversed.     But  we  attribute  to 
a  surface  the  amount  of  light  entering  the  eye  in  the  direction 
from  it  to  the  eye.     Hence  the  most  remote  portions  of  an  illu- 
minated surface  are  the  darkest  in  appearance. 

b.  Appearance  of  a  surface  of  shade  seen  obliquely.     A  surface  of 
shade  only  remits  to  the  eye  tertiary  rays,  as  they  may  be  called, 
that  is,  rays  received  by  reflection  from  surrounding  objects. 
Hence  the  atmospheric  reflections  coming  to  the  eye  in  the  Direc- 
tion of  the  shaded  object,  may  be  supposed  to  be  stronger  than 
the  light  from  the  shaded  object  itself.     But,  as  before,  we  attri- 
bute to  the  object  the  light  coming  to  the  eye  in  the  direction  of 
it,  while  the  more  remote  it  is,  the  greater  is  the  body  of  inter- 
vening air  which  sends  light  from  its  own  molecules  to  the  eye. 
Hence  the  more  remote  a  surface  of  shade  is,  the  lighter  it 
appears,  or  the  nearer  it  appears  to  be  of  the  same  brilliancy  of 
the  atmosphere  generally.     This  is  very  evident  in  nature,  when 
we  compare  the  lights  and  shades  of  distant  buildings,  or  hills, 
with  those  of  near  ones. 

c.  ^Results  of  the  two  preceding  principles.     1°.  A  series  of  illu- 
minated  parallel    surfaces,  seen    perpendicularly,   will   appear 
darker  and   darker,   as  they   become  more   distant.     2°.  The 
reverse  will  happen,  if  these  surfaces  are  in  shade.     3°.  Near 
shadows  will  appear  intense,  and  remote  ones  faint.    4°.  Portions 
of  illuminated  curved  surfaces,  which  are  more  distant  than  the 
brilliant  point,  or  line,  will  appear  darker  than  that  point  or  line. 
5°.  Portions  of  the  shade  of  curved  surfaces,  which  are  more 
distant  than  the  line  of  shade,  will  appear  lighter  than  the  line 
of  shade.     6°.  Curved  surfaces  will  appear  of  a  more  uniform 
tint,  the  more  distant  they  are,  the  lights  being  darkened,  accord- 
ing to  (a),  and  the  shades  dimmed,  according  to  (b). 

d.  Limitation  of  the  last  principle.     The  last  principle  applies, 
however,  to  the  relative  depths  of  a  body  of  air  of  finite  thickness, 
through  which  bodies  are  seen  from  an  infinite  distance.      See 
(119). 

e.  Effect  of  the  air  on  apparent  color.     The  blue  color  of  the  air, 


SHADES  AND  SHADOWS.  133 

due  to  the  rays  remitted  from  it  by  refraction,  as  explained  already, 
is  attributed  to  objects  seen  through  it,  as  well  as  the  intensity  o£ 
the  atmospheric  rays.  Hence  distant  objects  appear  of  a  bluish 
cast,  which  increases  in  clearness  with  their  distance.  It  is  thus 
that,  in  a  colored  drawing,  a  distant  object  may  be  distinguished 
from  the  dull  shading  due  to  uniformly  diffused  light  (120*7). 

123.  Effects  due  to  variations  in  the  intensity  of  the  light.  Inte- 
resting results  follow  an  examination  of  the  effect  of  lights,  each 
received  mainly  from  one  direction,  but  varying  in  intensity. 
All  so-called  transparent  media  absorb  some  light,  and,  for  the 
present  purpose,  it  is  sufficiently  correct  to  assume  that,  for  light 
passing  through  any  given  medium,  and  in  greater  quantity  than 
can  be  absorbed  by  the  medium,  the  absolute  amount  absorbed 
will  be  equal  for  all  degrees  of  intensity.  It  is  important  here 
to  observe  that  this  absorbed  light  is,  as  it  were,  latent,  and  does 
not  at  all  render  the  absorbing  body  visible. 

It  follows  that  in  case  of  a  feeble  light,  where  the  total  amount 
absorbed  by  the  air  and  by  surrounding  objects  is  large  in  com- 
parison to  the  whole  amount  of  light  emitted,  the  diffused  light 
caused  by  reflections  and  refractions  among  the  particles  of  air, 
and  the  light  remitted  from  surrounding  objects,  will  both  be 
small,  but  very  little  light  will  be  thrown  upon  the  shade,  or 
into  the  indefinite  shadow  in  space  of  a  body,  under  these  cir- 
cumstances, and  hence  this  shade  and  shadow  will  approach  in 
darkness  to  those  formed  under  the  supposition  that  they  received 
no  light,  and  were  absolutely  black  (105). 

Moreover,  the  diffused  light  spoken  of,  partially  illumines 
objects  within  the  indefinite  shadow  ;  but  if  it  be  feeble,  it  will 
all  be  absorbed  by  those  objects,  and  they  will  therefore,  as 
shown  above,  not  throw  any  light  into  the  indefinite  shadow,  or 
upon  the  shade  of  the  body.  For  this  additional  reason,  there- 
fore, will  the  shades  and  shadows  due  to  a  feeble  light  be 
intensely  dark,  as  compared  with  the  illumined  part  of  the  body. 
This  conclusion  corresponds  to  the  great  blackness  of  the  shades 
and  shadows  observable  on  and  about  buildings  by  moonlight. 
To  return,  now,  to  the  case  of  clear  sunlight,  and  then  to  apply 
these  principles,  it  appears  that,  while  the  absolute  amount  of 
absorption  remains  the  same,  there  is  a  large  excess  of  light, 
available  for  producing  diffused  and  reflected  light,  of  sufficient 
brightness  to  produce  reflection  from  bodies  in  the  shadow  upon 
the  shade  of  an  opaque  body,  so  that  enough  light  will  reach 


136  THEORY  OF  SHADING. 

this  shadow  and  shade,  to  mitigate  their  blackness    considera- 
bly. 

124.  Effects  due  to  the  forms  of  bodies. 

a.  Form  of  the  curves  of  equal  shade.     The  lines  of  equal  tint 
on  different  bodies,  will  be  either  similar  or  dissimilar.     They 
will  be  similar  on  developable  surfaces,  where  they  are  simply 
rectilinear  elements,  and  on  spheres,  ellipsoids,  etc.,  where  they 
will  generally  be  circles,  ellipses,  etc.     But  they  will  be  dissimi- 
lar on  the  more  complex  surfaces,  such  as  the  torus  and  pie- 
douche.     In  the  latter  case,  they  will  be  gradually  transformed, 
from  a  ring  surrounding  the  brilliant  point,  to  curves  more  and 
more  nearly  similar  to  the  curve  of  shade.     This  should  be 
remembered  in  distributing  the  tints  upon  such  surfaces. 

b.  Distribution  of  points  of  equal  shade.     Points  of  equal  ap- 
parent brilliancy  are  not,  however,  symmetrically  placed  around 
the  brilliant  point  as  a  centre.     That  is.  on  a  sphere,  for  example, 
the  curves  of  equal  illumination  are  not  circles,  having  their 
centres  in  the  diameter  through  the  brilliant  point.     See  PI. 
XV.,  Fig.  65,  which  represents  a  sphere,  with  the  plane  of  the 
paper  taken  as  a  plane  of  incident  and  reflected  rays,  as  LO  and 
OR/.     Then  B,  the  middle  point  of  the  arc  AC,  is  the  brilliant 
point. 

Now  from  (120a)  the  small  superficial  element,  shown  as 
having  A  for  its  centre,  is  the  element  having  the  greatest  actual 
illumination.  This  element  returns  the  broad  beam  of  rays, 
LA,  partly  by  reflection  (106)  in  the  condensed  beam  AR.  An 
equal  element,  equidistant  from  the  brilliant  point,  as  at  C,  re- 
ceives only  the  narrow  beam  of  rays  L/C,  which,  furthermore, 
it  remits  in  the  diffuse  beam  CR'.  It  is  now  quite  evident,  from 
a  comparison  of  the  breadth  of  the  two  reflected  beams,  that,  if 
they  diffused  equal  amounts  of  received  light,  the  intensity  of 
this  diffused  light,  and  consequently  the  apparent  brilliancy  of 
the  elements  A  and  C,  would  be  inversely  as  the  sine  of  the  angle 
made  by  the  reflected  ray  with  a  tangent  plane  at  the  point  from  which 
it  proceeds. 

But  not  only  is  the  beam  CR'  more  diluted,  so  to  speak,  than 
AR,  so  far  as  the  space  filled  by  a  given  amount  of  light  is  con- 
cerned, but  it  actually  contains  fewer  rays,  viz.,  those  of  the 
narrow  incident  beam  L'G.  Hence,  much  more,  is  the  element 
at  A,  brighter  than  the  element  C,  which  is  at  the  same  distance 
from  the  brilliant  point,  B. 


SHADES  AND  SHADOWS.  137 

In  fact,  the  superior  brilliancy  of  B  over  all  other  points,  can 
be  fully  accounted  for,  in  this  view,  only  by  taking  account  of 
the  effect  of  distance  (122a),  and  by  supposing  that  some  of  the 
reflecting  facets  of  molecules  in  the  element  A,  which  is  seen 
obliquely,  are  hidden  by  others  in  advance  of  them,  so  that  all 
the  rays  of  the  beam  LA  are  not  visibly  reflected. 

Therefore,  finally,  from  the  above,  we  conclude,  that  on  cylin- 
ders, cones,  spheres,  etc.,  points  which  are  equidistant  from  the 
brilliant  line,  or  point,  should  not  receive  equal  tints;  and  ob- 
serving, in  Fig.  63,  that  tangent  rays  parallel  to  LA  determine 
the  curve  of  shade,  that  the  darkest  of  two  such  points  should 
be  the  one  which  is  between  the  brilliant  element  and  the  Visible 
projection  of  the  element,  or  curve,  of  shade. 

c.  Relative  illumination  of  apparent  contours.     It  may  be  sepa- 
rately noted  that,  as  a  sort  of  balancing  between  the  results  of 
the  light  received  by  a  surface,  the  light  reflected  and  refracted 
from  it,  and  the  obstruction  of  some  of  these  remissions,  there 
will  be  a  narrow  space  along  the  apparent  contour  of  curved 
surfaces,  which  will  appear  lighter  than  the  portions  a  trifle 
further  from  that  contour;  for  this  space,  which  is  very  narrow 
in  projection,  is  wide  in  reality,  and  so  receives  a  broad  beam 
of  light ;  which,  however,  it  reflects  within  a  narrow  compass. 

d.  Rate  of  variation  of  shades  on  different  surfaces.     A  consi- 
deration of  the  forms  of  bodies  shows  that  there  must  be  varia- 
tions in  the  rate  of  gradation  from  light  to  dark,  on  different 
surfaces.     For  example,  see  PI.  XV.,  Figs.  59  and  60,  equal 
vertical  strips  of  shade  on  the  cylindrical  surface  will  not  appear 
equal  in  vertical  projection,  and  the  shading,  used  to  express  the 
curvature  of  its  convex  surface,  would  therefore  proceed  rapidly 
from  dark  to  medium  shades,  and  then,  more  gradually,  from 
medium  to  light  shades.     But  on  the  vertical  plane  seen  oblique- 
ly, Fig.  58,  similar  equal  strips  would  appear  equal  in  projection, 
and  the  shading  used  to  express  the  different  distances  of  these 
strips  from  the  eye  (122),  would  proceed  at  a  perfectly  uniform- 
rate  from  dark  to  light. 

e.  The  proper  shading  of  bounding  edges.     The  theory  of  the 
form  of  bounding  edges,  should  be  observed  in  the  execution  of 
shading.     These  edges  are  supposed  not  to  be  mathematical 
lines,  but  to  be  rounded,  through  the  ultimate  imperfections  of 
workmanship,  and  the  attrition  of  flying  particles.     Thus,  an 
edge  of  a  prism  is  to  be  regarded  as  minutely  cylindrical ;  and 


138  THEORY  OF  SHADING. 

the  circumference  of  the  base  of  a  cylinder  is  a  portion  of  a 
torus,  whose  meridian  section  is  a  very  minute  quadrant. 

These  edges,  even  when  in  shade,  will  have  their  own  brilliant 
points,  due  to  the  scattering  light  which  they  may  receive  from 
surrounding  objects,  and  to  their  superior  smoothness,  consequent 
on  greater  exposure  to  polishing  agencies.  Accordingly,  very 
pleasing  effects  are  produced,  especially  in  an  assemblage  of 
shaded  figures,  as  in  a  machine  drawing,  by  shading  the  edges 
of  each  form,  according  to  the  principles  here  stated. 

/.  Illustrations.  First.  On  a  shaded  drawing  of  a  vertical 
cylinder  of  revolution,  the  upper  edge  should  everywhere  be 
lighter  than  the  bordering  portions  of  the  convex  surface.  The 
lower  edge  of  its  illuminated  portion  would  be  darker  than  the 
convex  surface,  but  the  lower  edge  of  its  shade  should  be  lighter 
than  the  contiguous  shade,  though  rather  darker  than  the  upper 
edge  of  the  shade. 

Second.  The  successive  parallel  plane  surfaces,  forming  the 
front  of  an  edifice,  are  distinctly  shown  in  projection,  by  thus 
treating  their  edges,  which  are  mainly  horizontal  and  vertical 
quarter  cylinders.  Upper  and  left  hand  edges,  when  exposed  to 
the  light,  are  left  blank,  or  of  a  light  shade,  for  a  very  narrow 
space.  Lower  and  right  hand  edges,  are  quarter  cylinders  con- 
taining elements  of  shade,  and  may  be  ruled  with  a  tint  a  little 
darker  than  that  of  the  flat  surface  bounded  by  them. 

The  light  edges  may  be  ruled  with  white,  if  not  left  blank  in 
tinting.  This  course  would  produce  the  most  striking  effect  in 
drawings  on  tinted  paper.  By  thus  distinguishing  parallel  sur- 
faces by  the  treatment  of  their  edges,  we  avoid  the  necessity  of 
an  undue  exaggeration  of  the  difference  between  their  shades,  as 
explained  in  (122  c). 

125.  Effects  (on  the  drawing)  due  to  drawing  materials  and 
processes  of  manipulation. 

a.  Since  the  amount  of  remitted  light  received  by  the  eye, 
both  white,  or  reflected,  and  colored,  or  refracted,  decreases  in 
receding  from  the  brilliant  point,  while  the  number  of  colored 
rays  determines  the  apparent  clearness  and  strength  of  color,  it 
would  seem  that  the  projection  of  a  body  would  be  more  accu- 
rately shaded  by  a  mixture  of  India  ink  with  its  conventional  tint, 
than  by  shading  it  with  ink  alone,  to  express  its  gradations  of 
shades,  and  then  covering  it  with  a  flat  tint  of  the  conventional 
color.  For  the  latter  method  would  give  so  little  color  compara- 


SHADES  ANE   SHADOWS.  139 

lively  to  the  very  dark  shades,  that  it  would  be  imperceptible, 
while  in  fact  every  part  of  the  body  does  reveal  its  proper  color 
to  some  extent.  Otherwise,  if  the  conventional  flat  tint  were 
made  strong  enough  to  show,  at  the  line  of  shade,  it  would  obli- 
terate the  light  ink-shading  beneath  it,  in  the  vicinity  of  the  bril- 
liant point.  In  geometrical  drawing,  however,  one  of  the  exag- 
gerations, which  compensate  for  the  artificial  character  of  all 
projections,  is  secured  by  the  second  method  of  treating  colored 
objects.  Hence  the  view  presented  in  the  "  Elementary  Projec- 
tion Drawing"  (182). 

b.  There  are  three  simple  methods  which  may  be  used,  sepa- 
rately or  together,  in  executing  a  shaded  drawing.       These 
methods  are :  1.  By  softened  tints.     2.  By  superposed  flat  tints. 
3.  By  touches. 

c.  Shading  in  softened  tints.     This  is  difficult  to  execute,  but 
would  give,  when  perfectly  done,  the  most  perfectly  smooth  and 
continuous  gradation  of  shade.    Moreover,  it  requires  to  be  done 
at  one  operation,  with  a  mixture  of  ink  and  the  appropriate  con- 
ventional tint,  as  explained  in  (a).     For  if  the  two  shades  be 
made  separately,  though  both  by  this  same  method  of  softened 
tints,  discrepancies  between  their  gradations  will  be  apt  to  spoil 
the  intended  effect  of  each. 

d.  Shading  by  superposed  flat  tints.     This  consists  in  placing 
a  narrow  dark  tint  on  the  line  of  shade ;  then,  when  dry,  an- 
other, covering  the  first,  and  extending  a  little  further,  and  so 
on.     In  this  method,  the  drier  the  brush,  the  less  will  the  edges 
of  the  successive  bands  show. 

In  the  last  method,  we  rely  somewhat  on  a  free  conformity 
of  the  softened  shades  to  their  theoretically  true  distribution,  as 
determined  by  observation  merely. 

In  this  method,  we  aim  to  have  each  of  the  well  marked  suc- 
cessive bands  of  tint  exactly  located  so  as  to  represent  the  bands 
of  equal  apparent  brilliancy  on  the  body  represented.  They 
should  therefore,  on  account  of  their  conspicuousness,  be  located 
according  to  precise  knowledge  of  their  proper  position. 

But  this  location  it  is  difficult,  if  not  impossible,  to  effect,  by 
any  simple  or  generally  available  construction,  owing  to  the 
complication  of  the  problem,  even  in  the  simplest  cases,  as  indi- 
cated in  (124  I). 

e.  Shading  by  touches.     This  method  consists  in   placing 
directly  upon  each  part  of  the  drawing,  a  small  quantity  of  ink, 


140  THEORY  OF  SHADING. 

with  a  tolerably  dry  brush,  and  of  a  darkness  equal  to  that  of 
the  corresponding  point  of  the  surface  represented.  This  method 
is  quite  similar  to  that  of  shading  with  a  lead  pencil,  and, 
guided  by  practised  judgment,  will  secure  an  orderly  gradation 
of  shades,  without  involving  any  of  the  raggedness  and  blotted 
appearance  which  are  apt  to  appear  in  unskilful  use  of  the  two 
former  methods. 

/.  The  first  method  is  modified,  by  the  size  of  the  surface  of 
paper  to  be  shaded,  into  what  may  be  called  the  methods  by 
wet  softened  tints,  and  by  dry  softened  tints.  The  former  is 
applicable  to  quite  large  surfaces,  and  requires  a  pretty  large 
and  wet  brush.  The  latter  is  merely  the  second  method  with 
the  edge  of  each  tint  softened  by  a  brush  damp  with  clear  water. 

Neither  the  first  nor  the  second  method,  alone,  will  ever  pro- 
duce perfectly  smooth  shading.  The  third  must  be  combined 
with  them,  so  far  as  is  necessary  to  fill  up  the  irregularities  of 
shade  produced  by  those  preliminary  methods.  The  method  by 
touches  alone  is  very  tedious,  as  applied  to  large  surfaces.  Hence 
it  should  then  be  combined  with  the  second  method.  It  is, 
however,  conveniently  applied,  alone,  to  small  surfaces. 

g.  The  great  advantage  of  the  method  ty  touches,  is,  that  the 
minute  light  spots,  finally  left  among  the  fine  strokes  of  the 
brush,  reproduce,  in  the  drawing,  the  effect  of  the  minute  bril- 
liant points  of  the  asperities  of  the  object,  and  thus  give  the 
drawing  an  appearance  of  distinctness  and  solidity,  which  is 
very  desirable.  On  the  other  hand,  the  uniform  saturation  of 
the  paper  with  wet  tints,  used  in  the  method  of  softened  tints, 
produces  a  cloudy  effect  like  that  of  the  shade  of  polished 
bodies.  Yet  it  does  not,  on  the  whole,  adequately  represent 
such  bodies,  because  its  gradation  of  shade  is  gradual,  while 
theirs  is  apparently  abrupt,  on  account  of  the  dazzling  bright- 
ness of  their  brilliant  points. 

h.  The  effect  of  the  minute  brilliant  points  is  also  produced, 
in  part,  by  shading  on  rough  paper,  on  whose  actual  asperities 
the  light  may  act. 

i.  In  the  use  of  tinted  paper,  brilliant  effects  may  be  produced 
by  using  white  tints  for  the  brilliant  points. 

Also,  the  effect  of  colored  drawings  will  be  heightened,  by 
employing  tinted  paper  of  a  color  complementary  to  that  which 
prevails  on  the  drawing. 


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